- Use related multiplication facts. - Compare the place value of \(80\) and \(800\). - Consider how a larger divisor affects the quotient.

1. a) \(80 \div 2 = 40\). 2. b) \(80 \div 4 = 20\). 3. c) \(80 \div 8 = 10\). 4. d) \(800 \div 2 = 400\). 5. e) \(800 \div 8 = 100\).

a) \(40\) b) \(20\) c) \(10\) d) \(400\) e) \(100\)

- Use a related basic division fact for the leading digits. - Think of each dividend as a number of tens. - Ask how many times the divisor fits into the leading part of the dividend. - Check with multiplication.

1. \(120 \div 2 = 60\). 2. \(210 \div 3 = 70\). 3. \(320 \div 4 = 80\). 4. \(400 \div 8 = 50\). 5. \(540 \div 9 = 60\). 6. \(480 \div 6 = 80\).

\(60\), \(70\), \(80\), \(50\), \(60\), \(80\)

- Only the original value \(54\) matters. - Find one-half by dividing by \(2\). - Check by doubling your result.

1. One-half means divide by \(2\). 2. \(54\div2=27\). 3. The revised report should show \(27\).

\(27\)

- How many hundreds make \(600\)? - Share the hundreds first, using a familiar division fact. - Turn the number of hundreds back into blocks.

1. \(600\) is \(6\) hundreds. 2. Share \(6\) hundreds among \(3\) classes: each gets \(2\) hundreds. 3. So each class gets \(200\) blocks.

\(200\) blocks per class.

- Imagine sharing \(96\) counters equally into four groups. Does each part of the split belong to all four groups? - Think about how many groups of four are in \(80\), then in \(16\). - Check the result by multiplying the quotient by four.

1. \(80\div4=20\). 2. \(16\div4=4\). 3. Add the quotients: \(20+4=24\), so \(96\div4=24\).

\(80\div4=20\), \(16\div4=4\), and \(96\div4=24\).

- Look only at the hundreds before moving to the next column. - How many hundreds does each of three shares get? - Name the place value of that first quotient digit.

1. The first part is \(6\) hundreds divided into \(3\) equal shares. 2. Each share gets \(2\) hundreds, so the first quotient digit is \(2\) in the hundreds place.

\(2\) in the hundreds place.

- Try giving each student the same number of counters. - After making complete groups, count any counters still unused. - Ask whether every student could get one more.

1. Four equal groups of \(4\) counters use \(4\times4=16\) counters. 2. One counter is left, so the result is \(17\div4=4\) remainder \(1\). 3. The one leftover cannot give another counter to all four students.

\(4\) counters per student and \(1\) counter left over. One counter is not enough to give one more to each of the four students.

- Think about each place in the starting number separately. - If there is nothing to share in a place, what will each group receive there? - Check by multiplying the three-digit quotient by \(3\).

1. Each share gets \(6\div3=2\) hundreds. 2. There are no tens, so each share gets \(0\) tens. 3. Each share gets \(6\div3=2\) ones, so the quotient is \(202\).

\(606\div3=202\). The tens place is \(0\) because no tens need to be shared.

- For each dividend, find the greatest multiple of the divisor that does not exceed it. - The remainder must be less than the divisor. - Compare the three remainders only after all three divisions are complete.

1. a) \(47=6\times7+5\), so \(47\div6=7\) remainder \(5\). 2. b) \(60=7\times8+4\), so \(60\div7=8\) remainder \(4\). 3. c) \(85=9\times9+4\), so \(85\div9=9\) remainder \(4\). 4. Parts b and c have the same remainder, \(4\). 5. Part a has the greatest remainder, \(5\).

a) \(7\) remainder \(5\) b) \(8\) remainder \(4\) c) \(9\) remainder \(4\) d) b) and c) e) a), with remainder \(5\)

- Think about making equal groups of \(4\) apples. - Which multiple of \(4\) is closest to \(26\) without going over? - After making all the full groups, how many apples remain?

1. Divide the total number of apples by the number in each bag: \(26 \div 4\). 2. The greatest multiple of \(4\) that is no more than \(26\) is \(24\), because \(4 \times 6 = 24\). 3. Therefore, \(26 \div 4 = 6\) remainder \(2\). Lucas fills \(6\) bags, and \(2\) apples are left over.

Lucas can fill \(6\) bags completely. He will have \(2\) apples left over.

- Make equal groups of \(6\). - Which multiple of \(6\) is closest to \(57\) without going over? - What does the remainder represent in this situation?

1. Divide the number of pretzels by the number in each bag: \(57 \div 6\). 2. Since \(6 \times 9 = 54\) and \(57 - 54 = 3\), \(57 \div 6 = 9\) remainder \(3\). 3. The quotient gives the number of full bags, and the remainder gives the number of pretzels left over.

a) The baker can fill \(9\) bags completely. b) \(3\) pretzels will be left over.

- Divide from left to right, one place at a time. - Carry each remainder to the next place. - Write a zero in the quotient when the divisor does not fit into a place after a nonzero quotient digit has already been written.

1. For a), \(7 \div 6=1\) with remainder \(1\). Bring down the \(3\) to make \(13\). Then \(13 \div 6=2\) with remainder \(1\). Bring down the \(2\) to make \(12\), and \(12 \div 6=2\). Therefore, \(732 \div 6=122\). 2. For b), \(9 \div 4=2\) with remainder \(1\). Bring down the \(1\) to make \(11\). Then \(11 \div 4=2\) with remainder \(3\). Bring down the \(2\) to make \(32\), and \(32 \div 4=8\). Therefore, \(912 \div 4=228\). 3. For c), \(5 \div 5=1\). Bring down the \(2\). Since \(5\) goes into \(2\) zero times, write \(0\) in the tens place of the quotient. Bring down the final \(5\) to make \(25\), and \(25 \div 5=5\). Therefore, \(525 \div 5=105\).

a) \(122\) b) \(228\) c) \(105\)

- A remainder is the amount left when the divisor does not divide the dividend evenly. - The remainder must be less than the divisor. - Include a zero in the quotient when needed to preserve place value.

1. For a), \(8 \div 8=1\). Bring down the \(4\). Since \(8\) goes into \(4\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(9\) to make \(49\). Then \(49 \div 8=6\) with remainder \(1\). Therefore, \(849 \div 8=106\text{ R }1\). Check: \(8 \times 106+1=849\). 2. For b), begin with \(67\) because \(6<7\). Then \(67 \div 7=9\) with remainder \(4\). Bring down the final \(4\) to make \(44\). Then \(44 \div 7=6\) with remainder \(2\). Therefore, \(674 \div 7=96\text{ R }2\). Check: \(7 \times 96+2=674\).

a) \(106\text{ R }1\) b) \(96\text{ R }2\)

- If the first digit is less than the divisor, begin with the first two digits. - Divide from left to right. - Write zero in a quotient place when needed to preserve place value.

1. For a), begin with \(15\) because \(1<3\). Then \(15 \div 3=5\). Bring down the \(3\), and \(3 \div 3=1\). Bring down the \(6\), and \(6 \div 3=2\). Therefore, \(1536 \div 3=512\). 2. For b), \(24 \div 8=3\). Bring down the \(0\). Since \(8\) goes into \(0\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(8\), and \(8 \div 8=1\). Therefore, \(2408 \div 8=301\).

a) \(512\) b) \(301\)

- Divide from left to right. - Write zero in a quotient place when needed. - Multiply the quotient by the divisor to check.

1. For a), \(8 \div 6=1\) with remainder \(2\). Bring down the \(1\) to make \(21\), and \(21 \div 6=3\) with remainder \(3\). Bring down the \(6\) to make \(36\), and \(36 \div 6=6\). Therefore, \(816 \div 6=136\). Check: \(136 \times 6=816\). 2. For b), \(8 \div 4=2\). Bring down the \(1\). Since \(4\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 4=4\). Therefore, \(816 \div 4=204\). Check: \(204 \times 4=816\). 3. For c), \(8 \div 8=1\). Bring down the \(1\). Since \(8\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 8=2\). Therefore, \(816 \div 8=102\). Check: \(102 \times 8=816\).

a) \(136\); check: \(136 \times 6=816\) b) \(204\); check: \(204 \times 4=816\) c) \(102\); check: \(102 \times 8=816\)

- What total width is shared equally among the \(8\) boards? - Which operation finds the width of one equal share? - You can decompose the total into easier multiples of \(8\) before dividing.

1. Divide the total width by the number of boards: \(120 \div 8\). 2. Calculate: \(120 \div 8 = 15\). 3. Each board is \(15\,\text{cm}\) wide.

Each board is \(15\,\text{cm}\) wide.

- Think about how multiplication can help you make division equations with the same quotient. - Check whether each chosen one-digit divisor gives a dividend in the required range. - Use a different divisor each time.

1. Choose three different one-digit divisors whose products with \(90\) exceed \(200\), such as \(3\), \(4\), and \(5\). 2. Multiply each by \(90\): \(3\times90=270\), \(4\times90=360\), and \(5\times90=450\). 3. Reverse each multiplication fact: \(270\div3=90\), \(360\div4=90\), and \(450\div5=90\).

One possible set is \(270\div3=90\), \(360\div4=90\), and \(450\div5=90\).

- Look for a related basic division fact. - Use place value to account for the factors of \(10\). - Check each quotient with multiplication.

1. For a), use \(36 \div 4 = 9\), so \(360 \div 4 = 90\). 2. For b), use \(42 \div 6 = 7\), so \(420 \div 6 = 70\). 3. For c), use \(72 \div 9 = 8\), so \(720 \div 9 = 80\). 4. For d), use \(40 \div 8 = 5\), so \(4000 \div 8 = 500\).

a) \(90\) b) \(70\) c) \(80\) d) \(500\)

- Dividing finds the size of one equal addend. - Use the number of equal addends as the divisor. - Check by multiplying one addend by the number of addends.

1. \(8000\div5=1600\). 2. \(8000\div8=1000\). 3. \(8000\div4=2000\).

a) \(1600\) b) \(1000\) c) \(2000\)

- Equal sharing calls for division. - Use the number of warehouses as the divisor. - Check a share by multiplying it by the number of warehouses.

1. \(9600\div2=4800\). 2. \(9600\div4=2400\). 3. \(9600\div5=1920\). 4. \(9600\div8=1200\).

a) \(4800\) screws b) \(2400\) screws c) \(1920\) screws d) \(1200\) screws

- Find the one factor that combines with the given factor to make each column heading. - Use inverse multiplication or one-digit division. - Look for scaling relationships across a row.

1. For \(4000\), the missing factors are \(2000,1000,800,500\) for given factors \(2,4,5,8\). 2. For \(6000\), the missing factors are \(3000,1500,1200,750\). 3. For \(8000\), the missing factors are \(4000,2000,1600,1000\).

<table> <tr><th>Given factor</th><th>\(4000\)</th><th>\(6000\)</th><th>\(8000\)</th></tr> <tr><td>\(2\times\square\)</td><td>\(2000\)</td><td>\(3000\)</td><td>\(4000\)</td></tr> <tr><td>\(4\times\square\)</td><td>\(1000\)</td><td>\(1500\)</td><td>\(2000\)</td></tr> <tr><td>\(5\times\square\)</td><td>\(800\)</td><td>\(1200\)</td><td>\(1600\)</td></tr> <tr><td>\(8\times\square\)</td><td>\(500\)</td><td>\(750\)</td><td>\(1000\)</td></tr> </table>

- Divide by \(2\) to find one half. - Use basic halving facts and place value. - Check each result by doubling it.

1. \(3400 \div 2 = 1700\). 2. \(5600 \div 2 = 2800\). 3. \(1900 \div 2 = 950\). 4. \(7200 \div 2 = 3600\).

a) \(1700\) b) \(2800\) c) \(950\) d) \(3600\)

- Which numbers from a), b), and c) combine to make \(5228\)? - Divide each convenient part separately. - Add the partial quotients to find the final quotient.

1. \(4000 \div 4 = 1000\). 2. \(1200 \div 4 = 300\). 3. \(28 \div 4 = 7\). 4. Since \(5228 = 4000 + 1200 + 28\), add the partial quotients: \(1000 + 300 + 7 = 1307\).

a) \(1000\) b) \(300\) c) \(7\) d) \(1307\)

- For the weight estimate, find a nearby total that is easy to divide by \(5\). - For the price, work in cents so you can divide whole numbers. - What does “each peach costs the same” let you do?

1. Round the total weight of \(26\,\text{oz}\) to \(25\,\text{oz}\). Since \(25\div5=5\), the average weight is about \(5\,\text{oz}\). 2. \(\$2.45\) is \(245\) cents. Each peach has the same price, so \(245\div5=49\) cents.

a) About \(5\,\text{oz}\). b) \(49\) cents.

- Use the identical packages to work out the weight of one. - Convert the total shipping cost to cents. - What does the same cost for every package tell you about the second division?

1. The packages have equal weights: \(32\div4=8\), so each weighs \(8\,\text{lb}\). 2. Convert the total to cents: \(\$11.60=1160\) cents. Then \(1160\div4=290\) cents per package.

a) \(8\,\text{lb}\). b) \(290\) cents.

- Estimate the quotient first. - Divide from left to right and carry each remainder to the next place. - Check your answer with multiplication.

1. Divide the total amount by the number of teams: \(7536 \div 6\). 2. Using the standard division algorithm gives \(1256\). 3. Check: \(1256 \times 6=7536\).

Each team receives \(\$1256\).

- Use multiplication to check a quotient. - When there is a remainder, add it after multiplying the quotient and divisor. - Keep the division steps aligned by place value.

1. For a), begin with \(54\) because \(5<8\). Then \(54 \div 8=6\) with remainder \(6\). Bring down the \(3\) to make \(63\), and \(63 \div 8=7\) with remainder \(7\). Bring down the \(2\) to make \(72\), and \(72 \div 8=9\). Therefore, \(5432 \div 8=679\). Check: \(679 \times 8=5432\). 2. For b), \(7 \div 4=1\) with remainder \(3\). Bring down the \(8\) to make \(38\), and \(38 \div 4=9\) with remainder \(2\). Bring down the \(1\) to make \(21\), and \(21 \div 4=5\) with remainder \(1\). Bring down the \(5\) to make \(15\), and \(15 \div 4=3\) with remainder \(3\). Therefore, \(7815 \div 4=1953\text{ R }3\). Check: \(1953 \times 4+3=7815\).

a) \(679\); check: \(679 \times 8=5432\) b) \(1953\text{ R }3\); check: \(1953 \times 4+3=7815\)

- Divide one place at a time from left to right. - Carry each remainder to the next digit. - The final remainder must be less than the divisor.

1. For a), begin with \(39\) because \(3<7\). Then \(39 \div 7=5\) with remainder \(4\). Bring down the \(4\) to make \(44\). Then \(44 \div 7=6\) with remainder \(2\). Bring down the \(9\) to make \(29\). Then \(29 \div 7=4\) with remainder \(1\). Therefore, \(3949 \div 7=564\text{ R }1\). 2. For b), begin with \(25\) because \(2<4\). Then \(25 \div 4=6\) with remainder \(1\). Bring down the \(0\) to make \(10\). Then \(10 \div 4=2\) with remainder \(2\). Bring down the \(6\) to make \(26\). Then \(26 \div 4=6\) with remainder \(2\). Therefore, \(2506 \div 4=626\text{ R }2\). 3. For c), begin with \(61\) because \(6<8\). Then \(61 \div 8=7\) with remainder \(5\). Bring down the \(2\) to make \(52\). Then \(52 \div 8=6\) with remainder \(4\). Bring down the \(3\) to make \(43\). Then \(43 \div 8=5\) with remainder \(3\). Therefore, \(6123 \div 8=765\text{ R }3\).

a) \(564\text{ R }1\) b) \(626\text{ R }2\) c) \(765\text{ R }3\)

- Decide whether to use mental math or write intermediate steps for each division. - Divisibility rules can help you predict whether a remainder is possible. - Use the digit sum to check the division by \(9\).

1. \(1680 \div 2 = 840\). 2. \(1680 \div 3 = 560\). 3. \(1680 \div 4 = 420\). 4. \(1680 \div 5 = 336\). 5. \(1680 \div 6 = 280\). 6. \(1680 \div 7 = 240\). 7. \(1680 \div 8 = 210\). 8. \(1680 \div 9 = 186\) remainder \(6\).

\(1680 \div 2 = 840\); \(1680 \div 3 = 560\); \(1680 \div 4 = 420\); \(1680 \div 5 = 336\); \(1680 \div 6 = 280\); \(1680 \div 7 = 240\); \(1680 \div 8 = 210\); \(1680 \div 9 = 186\) remainder \(6\).

- Break \(1260\) into parts that are easier to divide. - Use the divisibility rule for \(9\) to check one result. - For division by \(8\), check whether the quotient multiplied by \(8\), plus the remainder, equals \(1260\).

1. \(1260 \div 7 = 180\), so the division by \(7\) has no remainder. 2. \(1260 \div 8 = 157\) remainder \(4\). 3. \(1260 \div 9 = 140\), so the division by \(9\) has no remainder. 4. The divisions by \(7\) and \(9\) have no remainder.

a) \(180\) b) \(157\) remainder \(4\) c) \(140\) The divisions by \(7\) and \(9\) have no remainder.

- Every student needs a place to sleep. - Does the number of full tents include all \(34\) students? - Can a tent hold fewer than \(4\) students?

1. Divide the number of students by the number each tent holds: \(34 \div 4 = 8\) remainder \(2\). 2. Eight tents hold \(8 \times 4 = 32\) students. 3. The remaining \(2\) students still need a tent, so one more tent is required. 4. The group needs \(9\) tents in all.

The group needs \(9\) tents. The remainder of \(2\) means that two students need an additional tent.

- How much dough is needed for one roll? - Is the remainder enough to make another full-size roll? - Interpret the remainder in the bakery context.

1. Divide the total dough by the dough needed for one roll: \(80 \div 9 = 8\) remainder \(8\). 2. Check the amount used: \(8 \times 9\,\text{oz} = 72\,\text{oz}\). 3. The remaining dough is \(80\,\text{oz} - 72\,\text{oz} = 8\,\text{oz}\). 4. Since \(8\,\text{oz}\) is less than \(9\,\text{oz}\), it is not enough for another whole roll.

The baker can make \(8\) whole rolls, with \(8\,\text{oz}\) of dough left over.

- Find the greatest multiple of the divisor below each dividend. - Subtract to find the remainder. - Check that the remainder is less than the divisor.

1. For part a, \(4 \times 3 = 12\), and \(14 - 12 = 2\), so the result is \(4\) remainder \(2\). 2. For part b, \(5 \times 4 = 20\), and \(22 - 20 = 2\), so the result is \(5\) remainder \(2\). 3. For part c, \(6 \times 5 = 30\), and \(31 - 30 = 1\), so the result is \(6\) remainder \(1\). 4. For part d, \(7 \times 6 = 42\), and \(43 - 42 = 1\), so the result is \(7\) remainder \(1\). 5. For part e, \(7 \times 7 = 49\), and \(52 - 49 = 3\), so the result is \(7\) remainder \(3\).

a) \(4\) remainder \(2\) b) \(5\) remainder \(2\) c) \(6\) remainder \(1\) d) \(7\) remainder \(1\) e) \(7\) remainder \(3\)

- Find the greatest multiple of each divisor that is less than the dividend. - Subtract that multiple from the dividend to find the remainder. - Check that each remainder is less than its divisor.

1. For part a, \(6 \times 3 = 18\), and \(19 - 18 = 1\), so the result is \(6\) remainder \(1\). 2. For part b, \(6 \times 4 = 24\), and \(26 - 24 = 2\), so the result is \(6\) remainder \(2\). 3. For part c, \(6 \times 5 = 30\), and \(33 - 30 = 3\), so the result is \(6\) remainder \(3\). 4. For part d, \(7 \times 6 = 42\), and \(47 - 42 = 5\), so the result is \(7\) remainder \(5\). 5. For part e, \(7 \times 7 = 49\), and \(52 - 49 = 3\), so the result is \(7\) remainder \(3\).

a) \(6\) remainder \(1\) b) \(6\) remainder \(2\) c) \(6\) remainder \(3\) d) \(7\) remainder \(5\) e) \(7\) remainder \(3\)

- Choose a nearby multiple of the divisor. - Break the dividend into that multiple and a remaining part. - Divide both parts by the divisor. - Add the two partial quotients.

1. For part a, break \(52\) into \(40 + 12\). Then \(40 \div 4 = 10\) and \(12 \div 4 = 3\), so \(10 + 3 = 13\). 2. For part b, break \(78\) into \(60 + 18\). Then \(60 \div 3 = 20\) and \(18 \div 3 = 6\), so \(20 + 6 = 26\). 3. For part c, break \(96\) into \(60 + 36\). Then \(60 \div 6 = 10\) and \(36 \div 6 = 6\), so \(10 + 6 = 16\). 4. For part d, break \(85\) into \(50 + 35\). Then \(50 \div 5 = 10\) and \(35 \div 5 = 7\), so \(10 + 7 = 17\).

a) \(13\) b) \(26\) c) \(16\) d) \(17\)

- Decompose each dividend into numbers that are easy to divide. - Use nearby multiples of the divisor. - Add the partial quotients.

1. Decompose \(92\) as \(80 + 12\): \(80 \div 4 = 20\) and \(12 \div 4 = 3\). Thus \(92 \div 4 = 23\). 2. Decompose \(75\) as \(60 + 15\): \(60 \div 3 = 20\) and \(15 \div 3 = 5\). Thus \(75 \div 3 = 25\).

There are \(23\) groups of \(4\) in \(92\), and \(25\) groups of \(3\) in \(75\).

- Choose two parts that are both multiples of the divisor. - Divide each part separately. - Add the partial quotients.

1. For part a, break \(168\) into \(150 + 18\). Then \(150 \div 3 = 50\) and \(18 \div 3 = 6\), so \(50 + 6 = 56\). 2. For part b, break \(455\) into \(420 + 35\). Then \(420 \div 7 = 60\) and \(35 \div 7 = 5\), so \(60 + 5 = 65\).

a) \(56\) b) \(65\)

- Break \(96\) into two numbers that are easy to divide by \(6\). - Divide each part by \(6\). - Add the partial quotients.

1. Divide the total number of tennis balls by the number in each basket: \(96 \div 6\). 2. Break \(96\) into compatible numbers: \(96 = 60 + 36\). 3. Divide each part: \(60 \div 6 = 10\) and \(36 \div 6 = 6\). 4. Add the partial quotients: \(10 + 6 = 16\).

The coach can fill \(16\) baskets completely.

- Break \(144\) into two numbers that are easy to divide by \(6\). - Divide each part by \(6\). - Add the partial quotients.

1. Divide the total number of rolls by the number of baskets: \(144 \div 6\). 2. Break \(144\) into compatible numbers: \(144 = 120 + 24\). 3. Divide each part: \(120 \div 6 = 20\) and \(24 \div 6 = 4\). 4. Add the partial quotients: \(20 + 4 = 24\).

Each basket contains \(24\) rolls.

- What operation finds one-eighth of a number? - Break \(864\) into two numbers that are easy to divide by \(8\). - Divide each part and combine the results.

1. To find one-eighth of \(864\), divide by \(8\): \(864 \div 8\). 2. Break \(864\) into compatible numbers: \(864 = 800 + 64\). 3. Divide each part: \(800 \div 8 = 100\) and \(64 \div 8 = 8\). 4. Add the partial quotients: \(100 + 8 = 108\).

The lifeboat model has \(108\) pieces.

- Which multiple of \(9\) is closest to \(87\) without going over? - How many groups of \(9\) does that multiple represent? - Subtract to find the remainder.

1. Divide the total number of students by the team size: \(87 \div 9\). 2. Since \(9 \times 9 = 81\) and \(87 - 81 = 6\), \(87 \div 9 = 9\) remainder \(6\). 3. The quotient is the number of complete teams, and the remainder is the number of students left over.

The students can form \(9\) complete teams, with \(6\) students left over.

- Find the greatest multiple of each divisor that does not exceed the dividend. - Subtract that multiple from the dividend to find the remainder. - Check that the remainder is less than the divisor.

1. For part a, \(7 \times 5 = 35\), and \(37 - 35 = 2\), so the result is \(7\) remainder \(2\). 2. For part b, \(7 \times 6 = 42\), and \(44 - 42 = 2\), so the result is \(7\) remainder \(2\). 3. For part c, \(8 \times 7 = 56\), and \(58 - 56 = 2\), so the result is \(8\) remainder \(2\). 4. For part d, \(8 \times 8 = 64\), and \(65 - 64 = 1\), so the result is \(8\) remainder \(1\). 5. For part e, \(8 \times 9 = 72\), and \(80 - 72 = 8\), so the result is \(8\) remainder \(8\).

a) \(7\) remainder \(2\) b) \(7\) remainder \(2\) c) \(8\) remainder \(2\) d) \(8\) remainder \(1\) e) \(8\) remainder \(8\)

- Divide the total amount of honey equally among \(5\) jars. - Use place value and a related basic division fact. - Include the unit in your answer.

1. Divide the total mass of honey by the number of jars: \(450 \div 5\). 2. Use \(45 \div 5 = 9\) and place-value reasoning to get \(450 \div 5 = 90\). 3. Each jar contains \(90\,\text{g}\) of honey.

Each jar contains \(90\,\text{g}\) of honey.

- Break \(252\) into two multiples of \(6\). - Divide both parts by \(6\). - Add the partial quotients.

1. Break \(252\) into \(240 + 12\), two numbers divisible by \(6\). 2. Divide each part: \(240 \div 6 = 40\) and \(12 \div 6 = 2\). 3. Add the partial quotients: \(40 + 2 = 42\).

Each planter gets \(42\) seedlings.

- The distance is shared equally across \(4\) hours. - Use place value and a related basic division fact. - State the distance traveled in one hour.

1. Divide the total distance by the number of hours: \(160 \div 4\). 2. Use \(16 \div 4 = 4\) and place-value reasoning to get \(160 \div 4 = 40\). 3. The cyclist averages \(40\) miles per hour.

The cyclist averages \(40\) miles per hour.

- Decompose each dividend into easy-to-divide parts. - Use related basic facts. - Check with multiplication.

1. a) Use \(42 \div 6 = 7\), so \(420 \div 6 = 70\). 2. b) Decompose \(560\): \(400 \div 4 = 100\) and \(160 \div 4 = 40\). Thus \(560 \div 4 = 140\). 3. c) Decompose \(750\): \(600 \div 3 = 200\) and \(150 \div 3 = 50\). Thus \(750 \div 3 = 250\).

a) \(70\) b) \(140\) c) \(250\)

- Convert the price to cents before dividing. - What operation finds half of an amount? - Convert the result back to dollars.

1. Write the book price in cents: \(\$6.40 = 640\text{ cents}\). 2. Divide by \(2\) to find half: \(640 \div 2 = 320\) cents. 3. Write \(320\) cents in dollars: \(320\text{ cents} = \$3.20\).

The magazine costs \(\$3.20\).

- Convert the total price to cents. - Divide the total cost equally among \(3\) notepads. - Convert the result back to dollars.

1. Write the total price in cents: \(\$7.20 = 720\text{ cents}\). 2. Divide the total price by the number of notepads: \(720 \div 3\). 3. Break \(720\) into compatible numbers: \(600 \div 3 = 200\) and \(120 \div 3 = 40\). Then add: \(200 + 40 = 240\) cents. 4. Convert \(240\) cents to dollars: \(240\text{ cents} = \$2.40\).

One notepad costs \(\$2.40\).

- Divide the total amount saved by the price of one shirt. - Use place value and a related basic division fact. - Check by multiplying the number of shirts by the price of one shirt.

1. Divide the total amount of money by the cost of each shirt: \(480 \div 8\). 2. Use \(48 \div 8 = 6\) and place-value reasoning to get \(480 \div 8 = 60\).

The club can buy \(60\) T-shirts.

- Find the greatest multiple of each divisor that does not exceed the dividend. - Subtract that multiple from the dividend to find the remainder. - Check that the remainder is less than the divisor.

1. For part a, \(9 \times 2 = 18\), and \(19 - 18 = 1\), so the result is \(9\) remainder \(1\). 2. For part b, \(8 \times 3 = 24\), and \(26 - 24 = 2\), so the result is \(8\) remainder \(2\). 3. For part c, \(9 \times 4 = 36\), and \(38 - 36 = 2\), so the result is \(9\) remainder \(2\). 4. For part d, \(9 \times 5 = 45\), and \(47 - 45 = 2\), so the result is \(9\) remainder \(2\). 5. For part e, \(9 \times 6 = 54\), and \(59 - 54 = 5\), so the result is \(9\) remainder \(5\).

a) \(9\) remainder \(1\) b) \(8\) remainder \(2\) c) \(9\) remainder \(2\) d) \(9\) remainder \(2\) e) \(9\) remainder \(5\)

- Find the greatest multiple of each divisor that does not exceed the dividend. - Use multiplication to determine the quotient. - Subtract to find the remainder.

1. a) \(8\times8=64\), so \(67\div8=8\) remainder \(3\). 2. b) \(9\times9=81\), so \(84\div9=9\) remainder \(3\). 3. c) \(7\times7=49\), so \(55\div7=7\) remainder \(6\). 4. d) \(8\times11=88\), so \(93\div8=11\) remainder \(5\). 5. e) \(9\times8=72\), so \(76\div9=8\) remainder \(4\).

a) \(8\) remainder \(3\) b) \(9\) remainder \(3\) c) \(7\) remainder \(6\) d) \(11\) remainder \(5\) e) \(8\) remainder \(4\)

- Find the greatest multiple of \(8\) that does not exceed each dividend. - Subtract that multiple to find the remainder. - Notice what happens to the remainder as the dividend increases by \(1\).

1. For part a, \(9 \times 8 = 72\), and \(78 - 72 = 6\), so the result is \(9\) remainder \(6\). 2. For part b, \(9 \times 8 = 72\), and \(79 - 72 = 7\), so the result is \(9\) remainder \(7\). 3. For part c, \(10 \times 8 = 80\), so the result is \(10\) remainder \(0\).

a) \(9\) remainder \(6\) b) \(9\) remainder \(7\) c) \(10\) remainder \(0\)

- Write zero in a quotient place when the divisor does not fit into that place. - Carry each remainder to the next digit. - The final remainder must be less than the divisor.

1. For a), \(8 \div 4=2\). Bring down the \(1\). Since \(4\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 4=4\). Therefore, \(816 \div 4=204\). 2. For b), \(7 \div 3=2\) with remainder \(1\). Bring down the \(2\) to make \(12\), and \(12 \div 3=4\). Bring down the \(5\), and \(5 \div 3=1\) with remainder \(2\). Therefore, \(725 \div 3=241\text{ R }2\). 3. For c), \(9 \div 5=1\) with remainder \(4\). Bring down the \(0\) to make \(40\), and \(40 \div 5=8\). Bring down the \(5\), and \(5 \div 5=1\). Therefore, \(905 \div 5=181\). 4. For d), \(6 \div 6=1\). Bring down the \(3\). Since \(6\) goes into \(3\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(8\) to make \(38\), and \(38 \div 6=6\) with remainder \(2\). Therefore, \(638 \div 6=106\text{ R }2\).

a) \(204\) b) \(241\text{ R }2\) c) \(181\) d) \(106\text{ R }2\)

- Find the greatest multiple of each divisor that does not exceed the dividend. - Subtract that multiple from the dividend to find the remainder. - Check that the remainder is less than the divisor.

1. For part a, \(8 \times 2 = 16\), and \(17 - 16 = 1\), so the result is \(8\) remainder \(1\). 2. For part b, \(7 \times 4 = 28\), and \(29 - 28 = 1\), so the result is \(7\) remainder \(1\). 3. For part c, \(8 \times 5 = 40\), and \(43 - 40 = 3\), so the result is \(8\) remainder \(3\). 4. For part d, \(6 \times 9 = 54\), and \(58 - 54 = 4\), so the result is \(6\) remainder \(4\). 5. For part e, \(8 \times 8 = 64\), and \(67 - 64 = 3\), so the result is \(8\) remainder \(3\).

a) \(8\) remainder \(1\) b) \(7\) remainder \(1\) c) \(8\) remainder \(3\) d) \(6\) remainder \(4\) e) \(8\) remainder \(3\)

- Break each dividend into two familiar multiples of the divisor. - Dividing ten groups of the divisor is often a useful first part. - Add the two partial quotients.

1. \(51 \div 3 = (30 \div 3) + (21 \div 3) = 10 + 7 = 17\). 2. \(68 \div 4 = (40 \div 4) + (28 \div 4) = 10 + 7 = 17\). 3. \(95 \div 5 = (50 \div 5) + (45 \div 5) = 10 + 9 = 19\). 4. \(84 \div 6 = (60 \div 6) + (24 \div 6) = 10 + 4 = 14\). 5. \(91 \div 7 = (70 \div 7) + (21 \div 7) = 10 + 3 = 13\). 6. \(104 \div 8 = (80 \div 8) + (24 \div 8) = 10 + 3 = 13\).

a) \(17\) b) \(17\) c) \(19\) d) \(14\) e) \(13\) f) \(13\)

- Check which two parts add up to each number being divided. - Each part can be divided by the same number as the whole. - Add the two partial quotients after you have calculated them.

1. The split \(70+42=112\) matches A, so A uses split 2. 2. Divide the two parts: \(70\div7=10\), \(42\div7=6\), and \(10+6=16\). 3. The split \(70+49=119\) matches B, so B uses split 1. 4. Divide the two parts: \(70\div7=10\), \(49\div7=7\), and \(10+7=17\).

A) Split 2; \(112\div7=16\). B) Split 1; \(119\div7=17\).

- Divide the total weight by the capacity of each crate. - Find \(84 \div 7\) for the large crates. - Find \(84 \div 4\) for the small crates.

1. For the large crates, divide by \(7\): \(84 \div 7 = 12\). 2. For the small crates, divide by \(4\): \(84 \div 4 = 21\).

Ms. Meyer would need \(12\) large crates or \(21\) small crates.

- Break \(72\) into two numbers that are easy to divide by \(4\). - Divide each part by \(4\). - Add the two partial quotients.

1. Divide the total number of chocolates by the number of children: \(72 \div 4\). 2. Break \(72\) into compatible numbers: \(72 = 40 + 32\). 3. Divide each part: \(40 \div 4 = 10\) and \(32 \div 4 = 8\). 4. Add the partial quotients: \(10 + 8 = 18\).

Each child gets \(18\) chocolates.

- Break \(156\) into two multiples of \(6\). - Divide each part by \(6\). - Add the partial quotients.

1. Divide the total number of seedlings by the number of rows: \(156 \div 6\). 2. Break \(156\) into compatible numbers: \(156 = 120 + 36\). 3. Divide each part: \(120 \div 6 = 20\) and \(36 \div 6 = 6\). 4. Add the partial quotients: \(20 + 6 = 26\).

There are \(26\) seedlings in each row.

- Divide the total number of pencils by each package size. - Use related basic division facts and place value. - Explain how a smaller package size affects the number of packages.

1. For packages of \(8\), divide: \(160 \div 8 = 20\). 2. For packages of \(4\), divide: \(160 \div 4 = 40\).

The store can fill \(20\) packages of \(8\) pencils or \(40\) packages of \(4\) pencils.

- Find the greatest multiple of each divisor that does not exceed the dividend. - Subtract that multiple from the dividend to find the remainder. - Use your multiplication facts to check each quotient.

1. For part a, \(6 \times 8 = 48\), and \(52 - 48 = 4\), so the result is \(6\) remainder \(4\). 2. For part b, \(7 \times 5 = 35\), and \(37 - 35 = 2\), so the result is \(7\) remainder \(2\). 3. For part c, \(9 \times 7 = 63\), and \(66 - 63 = 3\), so the result is \(9\) remainder \(3\). 4. For part d, \(7 \times 4 = 28\), and \(29 - 28 = 1\), so the result is \(7\) remainder \(1\). 5. For part e, \(9 \times 9 = 81\), and \(83 - 81 = 2\), so the result is \(9\) remainder \(2\).

a) \(6\) remainder \(4\) b) \(7\) remainder \(2\) c) \(9\) remainder \(3\) d) \(7\) remainder \(1\) e) \(9\) remainder \(2\)

- The dividend is the number being divided. - The divisor is the number you divide by. - Recall the term for the result of division.

1. Place the dividend before the division symbol and the divisor after it: \(126 \div 9\). 2. Evaluate: \(126 \div 9 = 14\). 3. The result of a division is called the quotient.

\(126 \div 9 = 14\). The result, \(14\), is the quotient.

- Find the greatest multiple of \(9\) that does not exceed \(76\). - Subtract that multiple from \(76\) to find the remainder. - Check by multiplying the quotient by \(9\) and adding the remainder.

1. The greatest multiple of \(9\) that does not exceed \(76\) is \(8 \times 9 = 72\). 2. The remainder is \(76 - 72 = 4\), so \(76 \div 9 = 8\) remainder \(4\). 3. Check: \(8 \times 9 + 4 = 72 + 4 = 76\).

\(76 \div 9 = 8\) remainder \(4\). Check: \(8 \times 9 + 4 = 76\).

- Break \(744\) into numbers that are easy to divide by \(6\). - Divide each part by \(6\). - Add the partial quotients and include the dollar sign.

1. Divide the total donation by the number of classes: \(744 \div 6\). 2. Decompose \(744\) into compatible numbers: \(744 = 600 + 120 + 24\). 3. Divide each part: \(600 \div 6 = 100\), \(120 \div 6 = 20\), and \(24 \div 6 = 4\). 4. Add the partial quotients: \(100 + 20 + 4 = 124\).

Each class receives \(\$124\).

- Compare the size of a remainder with the divisor. - Think about whether another group of \(4\) can be made from a remainder of \(5\). - List all whole numbers less than \(4\).

1. A remainder must always be less than the divisor. 2. Since \(5 > 4\), a remainder of \(5\) is not possible when dividing by \(4\). Another full group of \(4\) could be made. 3. The possible remainders are the whole numbers less than \(4\): \(0\), \(1\), \(2\), and \(3\).

Leo is mistaken because a remainder must be less than the divisor. When dividing by \(4\), the possible remainders are \(0\), \(1\), \(2\), and \(3\).

- Divide from left to right. - Write zero in a quotient place when the divisor does not fit into that place. - Check each quotient with multiplication.

1. For a), begin with \(45\) because \(4<5\). Then \(45 \div 5=9\). Bring down the \(0\), and write \(0\) in the tens place of the quotient. Bring down the \(5\), and \(5 \div 5=1\). Therefore, \(4505 \div 5=901\). 2. For b), \(12 \div 6=2\). Bring down the \(4\). Since \(6\) goes into \(4\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(8\) to make \(48\), and \(48 \div 6=8\). Therefore, \(1248 \div 6=208\). 3. For c), begin with \(56\) because \(5<7\). Then \(56 \div 7=8\). Bring down the \(8\), and \(8 \div 7=1\) with remainder \(1\). Bring down the \(4\) to make \(14\), and \(14 \div 7=2\). Therefore, \(5684 \div 7=812\). 4. For d), \(8 \div 8=1\). Bring down the \(1\). Since \(8\) goes into \(1\) zero times, write \(0\) in the hundreds place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 8=2\). Bring down the final \(8\), and \(8 \div 8=1\). Therefore, \(8168 \div 8=1021\).

a) \(901\) b) \(208\) c) \(812\) d) \(1021\)

- Divide each total number of posters by the number of hours. - The quotient represents posters per hour. - Compare the two quotient values.

1. Machine A: \(1386 \div 6=231\) posters per hour. 2. Machine B: \(1848 \div 8=231\) posters per hour. 3. The hourly rates are equal.

a) Machine A prints \(231\) posters per hour, and machine B also prints \(231\) posters per hour. b) The machines print at the same rate.

- Divide the total amount by the price per towel. - Repeat the division with the second unit price. - Check each quotient by multiplication.

1. At \(\$7\) each, the club buys \(\$1610 \div \$7 = 230\) towels. 2. At \(\$5\) each, the club could buy \(\$1610 \div \$5 = 322\) towels.

a) The club buys \(230\) towels. b) At the other supplier, it could buy \(322\) towels.

- Divide from left to right, one place at a time. - Carry each remainder to the next digit. - Check using multiplication and the remainder.

1. Begin with \(7\). Then \(7 \div 4=1\) with remainder \(3\). Bring down the \(4\) to make \(34\), and \(34 \div 4=8\) with remainder \(2\). Bring down the \(5\) to make \(25\), and \(25 \div 4=6\) with remainder \(1\). Bring down the \(2\) to make \(12\), and \(12 \div 4=3\). Therefore, \(7452 \div 4=1863\). 2. Begin with \(9\). Then \(9 \div 7=1\) with remainder \(2\). Bring down the \(6\) to make \(26\), and \(26 \div 7=3\) with remainder \(5\). Bring down the \(0\) to make \(50\), and \(50 \div 7=7\) with remainder \(1\). Bring down the \(5\) to make \(15\), and \(15 \div 7=2\) with remainder \(1\). Therefore, \(9605 \div 7=1372\text{ R }1\). 3. Begin with \(4\). Then \(4 \div 3=1\) with remainder \(1\). Bring down the \(8\) to make \(18\), and \(18 \div 3=6\). Bring down the \(1\). Since \(3\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(5\) to make \(15\), and \(15 \div 3=5\). Therefore, \(4815 \div 3=1605\). 4. Begin with \(5\). Since \(6\) does not fit into \(5\), use \(52\). Then \(52 \div 6=8\) with remainder \(4\). Bring down the \(3\) to make \(43\), and \(43 \div 6=7\) with remainder \(1\). Bring down the \(6\) to make \(16\), and \(16 \div 6=2\) with remainder \(4\). Therefore, \(5236 \div 6=872\text{ R }4\).

1. \(1863\) 2. \(1372\text{ R }1\) 3. \(1605\) 4. \(872\text{ R }4\)

- Divide from left to right. - Write zero in a quotient place when needed. - Check whether a final remainder is left and verify with multiplication.

1. For a), begin with \(43\) because \(4<7\). Then \(43 \div 7=6\) with remainder \(1\). Bring down the \(6\) to make \(16\), and \(16 \div 7=2\) with remainder \(2\). Bring down the \(8\) to make \(28\), and \(28 \div 7=4\). Therefore, \(4368 \div 7=624\). 2. For b), begin with \(29\) because \(2<4\). Then \(29 \div 4=7\) with remainder \(1\). Bring down the \(5\) to make \(15\), and \(15 \div 4=3\) with remainder \(3\). Bring down the \(2\) to make \(32\), and \(32 \div 4=8\). Therefore, \(2952 \div 4=738\). 3. For c), begin with \(50\) because \(5<8\). Then \(50 \div 8=6\) with remainder \(2\). Bring down the \(4\) to make \(24\), and \(24 \div 8=3\). Bring down the \(3\). Since \(8\) goes into \(3\) zero times, write \(0\) in the ones place of the quotient, with remainder \(3\). Therefore, \(5043 \div 8=630\text{ R }3\).

a) \(624\) b) \(738\) c) \(630\text{ R }3\)

- Write zero in a quotient place when the divisor fits zero times there. - A final remainder must be less than the divisor. - Check division by multiplying the quotient and divisor, then adding any remainder.

1. For a), begin with \(48\) because \(4<8\). Then \(48 \div 8=6\). Bring down the \(1\). Since \(8\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 8=2\). Therefore, \(4816 \div 8=602\). 2. For b), begin with \(36\) because \(3<4\). Then \(36 \div 4=9\). Bring down the \(0\), and write \(0\) in the tens place of the quotient. Bring down the \(8\), and \(8 \div 4=2\). Therefore, \(3608 \div 4=902\). 3. For c), begin with \(18\) because \(1<9\). Then \(18 \div 9=2\). Bring down the \(4\). Since \(9\) goes into \(4\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(8\) to make \(48\), and \(48 \div 9=5\) with remainder \(3\). Therefore, \(1848 \div 9=205\text{ R }3\). 4. For d), \(7 \div 7=1\). Bring down the \(2\). Since \(7\) goes into \(2\) zero times, write \(0\) in the hundreds place of the quotient. Bring down the \(1\) to make \(21\), and \(21 \div 7=3\). Bring down the final \(0\), and write \(0\) in the ones place of the quotient. Therefore, \(7210 \div 7=1030\). 5. Check for d): \(1030 \times 7=7210\).

a) \(602\) b) \(902\) c) \(205\text{ R }3\) d) \(1030\); check: \(1030 \times 7=7210\)

- Divide the total cost by the price of one kit. - Break \(168\) into multiples of \(7\). - Check the quotient by multiplication.

1. Divide the total cost by the cost of one art kit: \(168 \div 7\). 2. Break \(168\) into compatible numbers: \(168 = 140 + 28\). 3. Divide each part: \(140 \div 7 = 20\) and \(28 \div 7 = 4\). 4. Add the partial quotients: \(20 + 4 = 24\).

The school buys \(24\) art kits.

- Compare the first digit of each dividend with the divisor. - If the first digit is too small to divide by the divisor, division starts with the first two digits. - Use the starting place to decide how many quotient digits there will be; you do not need the exact quotient.

1. For a), the leading digit \(8\) is at least the divisor \(5\), so division begins in the thousands place. The quotient has \(4\) digits. 2. For b), the leading digit \(2\) is less than \(6\), so division begins with \(24\) in the first two places. The quotient has \(3\) digits. 3. For c), the leading digit \(3\) is less than \(9\), so division begins with \(31\) in the first two places. The quotient has \(3\) digits.

a) \(4\) digits b) \(3\) digits c) \(3\) digits

- Break \(255\) into two multiples of \(3\). - Divide both parts by \(3\). - Add the partial quotients.

1. Divide the total by \(3\): \(255 \div 3\). 2. Break \(255\) into \(240 + 15\). 3. Divide each part: \(240 \div 3 = 80\) and \(15 \div 3 = 5\). 4. Add the partial quotients: \(80 + 5 = 85\).

Each class receives \(\$85\).

- Solve two separate division problems. - Break \(432\) into compatible numbers for each divisor. - Check whether the larger case size gives more or fewer cases.

1. For cases of \(6\), divide: \(432 \div 6\). Break apart \(432 = 420 + 12\), so \(420 \div 6 = 70\) and \(12 \div 6 = 2\). Therefore, \(70 + 2 = 72\). 2. For cases of \(8\), divide: \(432 \div 8\). Break apart \(432 = 400 + 32\), so \(400 \div 8 = 50\) and \(32 \div 8 = 4\). Therefore, \(50 + 4 = 54\).

The company needs \(72\) cases that hold \(6\) bottles or \(54\) cases that hold \(8\) bottles.

- Divide \(360\) by \(6\) for part a. - For part b, compare the two carton capacities. - Think about how group size affects the number of groups when the total stays the same.

1. For part a, divide the number of jars by the number in each carton: \(360 \div 6 = 60\). 2. For part b, each larger carton holds more jars. Packing the same total into larger groups produces fewer groups, so fewer cartons are needed.

a) The beekeeper can fill \(60\) cartons. b) Fewer cartons are needed because each carton holds more jars.

- Break \(480\) into two numbers that are easy to divide by \(4\). - Divide each part by \(4\). - Add the partial quotients and include the unit.

1. Divide the total mass by the number of barrels: \(480 \div 4\). 2. Break \(480\) into compatible numbers: \(480 = 400 + 80\). 3. Divide each part: \(400 \div 4 = 100\) and \(80 \div 4 = 20\). 4. Add the partial quotients: \(100 + 20 = 120\).

Each barrel contains \(120\,\text{kg}\) of honey.

- Look carefully at whether the missing number is the dividend, divisor, or quotient. - A multiplication fact can help with a missing number in division. - Substitute your answer into each equation and check it.

1. In a), \(40\times3=120\), so the missing divisor is \(3\). 2. In b), \(40\times5=200\), so the missing dividend is \(200\). 3. In c), \(8\times36=288\), so the quotient is \(36\). 4. In d), \(9\times42=378\), so the quotient is \(42\). 5. In e), \(40\times2=80\), so the missing dividend is \(80\).

a) \(3\) b) \(200\) c) \(36\) d) \(42\) e) \(80\)

- Break each dividend into familiar multiples of the divisor. - Divide the parts separately, then add the partial quotients.

1. For part a, \(300 \div 3 = 100\) and \(60 \div 3 = 20\). Then \(100 + 20 = 120\). 2. For part b, \(400 \div 4 = 100\) and \(120 \div 4 = 30\). Then \(100 + 30 = 130\). 3. For part c, \(500 \div 5 = 100\) and \(250 \div 5 = 50\). Then \(100 + 50 = 150\).

a) \(120\) b) \(130\) c) \(150\)

- Divide the total number of marbles by \(6\). - Use place value and a related basic division fact. - Check your quotient by multiplication.

1. Divide the total number of marbles by the number in each bag: \(540 \div 6\). 2. Use \(54 \div 6 = 9\) and place-value reasoning to get \(540 \div 6 = 90\).

The store can fill \(90\) bags.

- Find the greatest multiple of each divisor that does not exceed the dividend. - Subtract to find the remainder. - Check with \(\text{quotient} \times \text{divisor} + \text{remainder} = \text{dividend}\).

1. For part a, \(7 \times 4 = 28\), and \(29 - 28 = 1\). Therefore, \(29 \div 4 = 7\) remainder \(1\). Check: \(7 \times 4 + 1 = 29\). 2. For part b, \(9 \times 5 = 45\), and \(46 - 45 = 1\). Therefore, \(46 \div 5 = 9\) remainder \(1\). Check: \(9 \times 5 + 1 = 46\). 3. For part c, \(7 \times 8 = 56\), and \(58 - 56 = 2\). Therefore, \(58 \div 8 = 7\) remainder \(2\). Check: \(7 \times 8 + 2 = 58\). 4. For part d, \(8 \times 9 = 72\), and \(74 - 72 = 2\). Therefore, \(74 \div 9 = 8\) remainder \(2\). Check: \(8 \times 9 + 2 = 74\).

a) \(7\) remainder \(1\); check: \(7 \times 4 + 1 = 29\) b) \(9\) remainder \(1\); check: \(9 \times 5 + 1 = 46\) c) \(7\) remainder \(2\); check: \(7 \times 8 + 2 = 58\) d) \(8\) remainder \(2\); check: \(8 \times 9 + 2 = 74\)

- Think of the missing factor as the number that makes \(8\times\square=320\). - Use division by the known factor to find the missing factor. - Check your quotient by multiplying it by \(8\).

1. Divide the product by the known factor: \(320\div8=40\). 2. Check: \(8\times40=320\).

The other factor is \(40\).

- Find the greatest multiple of \(9\) that does not exceed \(86\). - Subtract that multiple from \(86\). - The multiple tells you the quotient, and the difference is the remainder.

1. The greatest multiple of \(9\) that does not exceed \(86\) is \(9 \times 9 = 81\). 2. The remainder is \(86 - 81 = 5\). 3. Therefore, \(86 \div 9 = 9\) remainder \(5\).

\(86 \div 9 = 9\) remainder \(5\).

- Think about splitting all the pencils into equal groups. - Can you break the total into smaller amounts that are easier to divide? - For part b, compare how many pencils fit in one pack.

1. Divide the pencils into packs: \(825 \div 5\). 2. Split \(825 = 500 + 300 + 25\). Then \(500 \div 5 = 100\), \(300 \div 5 = 60\), and \(25 \div 5 = 5\). 3. Add: \(100 + 60 + 5 = 165\) full packs. 4. Packs that hold up to \(10\) pencils would hold more each, so fewer packs would be needed.

a) \(165\) full packs. b) Fewer packs, because each pack could hold more pencils.

- Compare each remainder with the divisor \(4\). - Think about whether a full group of \(4\) can be made from a remainder of \(4\). - List all whole numbers less than \(4\).

1. A remainder must be less than the divisor. 2. When dividing by \(4\), the only possible remainders are \(0\), \(1\), \(2\), and \(3\). 3. The listed remainder \(4\) is invalid because another full group of \(4\) could be made, leaving remainder \(0\).

The remainder \(4\) is incorrect. A remainder must be less than the divisor, so division by \(4\) can have only remainders \(0\), \(1\), \(2\), or \(3\).

- Break \(168\) into two multiples of \(6\). - Divide each part by \(6\). - Add the partial quotients.

1. Break \(168\) into \(120 + 48\), two numbers divisible by \(6\). 2. Divide each part: \(120 \div 6 = 20\) and \(48 \div 6 = 8\). 3. Add the partial quotients: \(20 + 8 = 28\).

Each child receives \(\$28\).

- Break \(161\) into two multiples of \(7\). - Divide each part by \(7\). - Add the partial quotients.

1. Divide the total number of bulbs by the number in each row: \(161 \div 7\). 2. Break \(161\) into compatible numbers: \(161 = 140 + 21\). 3. Divide each part: \(140 \div 7 = 20\) and \(21 \div 7 = 3\). 4. Add the partial quotients: \(20 + 3 = 23\).

The gardener can plant \(23\) complete rows.

- Choose two parts that are both multiples of the divisor. - Find the remaining part after choosing the first partial dividend. - Add the partial quotients.

1. For part a, \(72 = 60 + 12\). Then \((60 \div 3) + (12 \div 3) = 20 + 4 = 24\). 2. For part b, \(95 = 50 + 45\). Then \((50 \div 5) + (45 \div 5) = 10 + 9 = 19\). 3. For part c, \(84 = 80 + 4\). Then \((80 \div 4) + (4 \div 4) = 20 + 1 = 21\).

a) \(72 \div 3 = (60 \div 3) + (12 \div 3) = 20 + 4 = 24\) b) \(95 \div 5 = (50 \div 5) + (45 \div 5) = 10 + 9 = 19\) c) \(84 \div 4 = (80 \div 4) + (4 \div 4) = 20 + 1 = 21\)

- Decompose each dividend into place-value parts that are easy to divide. - Divide each part by the divisor. - Add the partial quotients.

1. a) Decompose \(480\) as \(400 + 80\): \(400 \div 2 = 200\) and \(80 \div 2 = 40\). The quotient is \(240\). 2. b) Decompose \(606\) as \(600 + 6\): \(600 \div 3 = 200\) and \(6 \div 3 = 2\). The quotient is \(202\). 3. c) Decompose \(909\) as \(900 + 9\): \(900 \div 9 = 100\) and \(9 \div 9 = 1\). The quotient is \(101\). 4. d) Decompose \(840\) as \(800 + 40\): \(800 \div 4 = 200\) and \(40 \div 4 = 10\). The quotient is \(210\). 5. e) Decompose \(550\) as \(500 + 50\): \(500 \div 5 = 100\) and \(50 \div 5 = 10\). The quotient is \(110\).

a) \(240\) b) \(202\) c) \(101\) d) \(210\) e) \(110\)

- Decide what operation helps when the dividend is missing. - Rewrite each division equation as multiplication. - Look for a related basic multiplication fact. - Check every completed equation.

1. a) Since \(20 \times 8 = 160\), the divisor is \(8\). 2. b) \(60 \times 5 = 300\), so the dividend is \(300\). 3. c) \(720 \div 8 = 90\). 4. d) Since \(90 \times 3 = 270\), the divisor is \(3\). 5. e) \(40 \times 7 = 280\), so the dividend is \(280\).

a) \(8\) b) \(300\) c) \(90\) d) \(3\) e) \(280\)

- Break each dividend into a large place-value number and a smaller multiple of the divisor. - Divide both parts. - Add the partial quotients.

1. For part a, \(428 = 400 + 28\). Then \(400 \div 4 = 100\) and \(28 \div 4 = 7\), so the quotient is \(107\). 2. For part b, \(618 = 600 + 18\). Then \(600 \div 6 = 100\) and \(18 \div 6 = 3\), so the quotient is \(103\). 3. For part c, \(927 = 900 + 27\). Then \(900 \div 9 = 100\) and \(27 \div 9 = 3\), so the quotient is \(103\). 4. For part d, \(515 = 500 + 15\). Then \(500 \div 5 = 100\) and \(15 \div 5 = 3\), so the quotient is \(103\).

a) \(107\) b) \(103\) c) \(103\) d) \(103\)

- Calculate all five quotients. - Look for the quotient that occurs most often. - Identify the single different result.

1. The quotients are \(240 \div 3 = 80\), \(400 \div 5 = 80\), \(560 \div 7 = 80\), \(210 \div 3 = 70\), and \(640 \div 8 = 80\). 2. Four expressions have quotient \(80\), while d) has quotient \(70\).

d) \(210 \div 3\) does not belong because its quotient is \(70\); all the others have quotient \(80\).

- Find a large familiar multiple of the divisor below the dividend. - Divide that part and the remaining part separately. - Add the partial quotients.

1. For part a, break \(378\) into \(360 + 18\). Then \(360 \div 6 = 60\) and \(18 \div 6 = 3\), so \(60 + 3 = 63\). 2. For part b, break \(532\) into \(490 + 42\). Then \(490 \div 7 = 70\) and \(42 \div 7 = 6\), so \(70 + 6 = 76\).

a) \(378 \div 6 = 63\) b) \(532 \div 7 = 76\)

- Break \(544\) into two multiples of \(8\). - Divide each part by \(8\). - Add the partial quotients.

1. Divide the total number of boxes by the number of carts: \(544 \div 8\). 2. Break \(544\) into compatible numbers: \(544 = 480 + 64\). 3. Divide each part: \(480 \div 8 = 60\) and \(64 \div 8 = 8\). 4. Add the partial quotients: \(60 + 8 = 68\).

Each rolling cart holds \(68\) boxes.

- Compare the divisors \(2\), \(4\), and \(8\). - Think about what happens to a quotient when the divisor becomes larger. - Use the first quotient to find the second, then the third. - Describe the inverse change between divisor and quotient.

1. \(640 \div 2 = 320\). 2. The divisor doubles from \(2\) to \(4\), so the quotient is halved: \(640 \div 4 = 160\). 3. The divisor doubles again from \(4\) to \(8\), so the quotient is halved again: \(640 \div 8 = 80\). 4. With a fixed dividend, doubling the divisor halves the quotient.

a) \(320\) b) \(160\) c) \(80\) When the divisor doubles, the quotient is halved.

- Break each dividend into two multiples of the divisor. - Start with a large multiple that is easy to divide. - Divide the remaining part separately. - Add the partial quotients.

1. For part a, \(84 = 60 + 24\). Then \(60 \div 3 = 20\) and \(24 \div 3 = 8\), so the quotient is \(28\). 2. For part b, \(75 = 50 + 25\). Then \(50 \div 5 = 10\) and \(25 \div 5 = 5\), so the quotient is \(15\). 3. For part c, \(92 = 80 + 12\). Then \(80 \div 4 = 20\) and \(12 \div 4 = 3\), so the quotient is \(23\). 4. For part d, \(108 = 60 + 48\). Then \(60 \div 6 = 10\) and \(48 \div 6 = 8\), so the quotient is \(18\).

a) \(28\) b) \(15\) c) \(23\) d) \(18\)

- Evaluate the left quotient first. - Evaluate the right quotient. - Decompose a dividend into easy-to-divide addends when helpful. - Compare the two quotients at the end.

1. a) \(91 \div 7 = 13\) and \(84 \div 6 = 14\), so \(13 < 14\). 2. b) \(112 \div 8 = 14\) and \(96 \div 4 = 24\), so \(14 < 24\).

a) \(<\) b) \(<\)

- Divide the large place-value part first. - Divide the remaining part by the same divisor. - Add the partial quotients.

1. For Problem 1, \(800 \div 4 = 200\) and \(24 \div 4 = 6\). Then \(200 + 6 = 206\). 2. For Problem 2, \(900 \div 9 = 100\) and \(18 \div 9 = 2\). Then \(100 + 2 = 102\).

Problem 1: \(800 \div 4 = 200\), \(24 \div 4 = 6\), quotient: \(206\) Problem 2: \(900 \div 9 = 100\), \(18 \div 9 = 2\), quotient: \(102\)

- Find a nearby multiple of the divisor with a zero in the ones place. - Break the dividend into that large part and a smaller remaining part. - Divide both parts separately. - Add the partial quotients.

1. For part a, \((140 \div 7) + (7 \div 7) = 20 + 1 = 21\). 2. For part b, \((250 \div 5) + (5 \div 5) = 50 + 1 = 51\). 3. For part c, \((320 \div 8) + (48 \div 8) = 40 + 6 = 46\). 4. For part d, \((420 \div 6) + (6 \div 6) = 70 + 1 = 71\).

a) \(21\) b) \(51\) c) \(46\) d) \(71\)

- Break each dividend into a hundreds part and a remaining part. - Divide the hundreds part by the divisor. - Divide the remaining part by the divisor. - Add the partial quotients.

1. For part a, \(424 = 400 + 24\). Then \(400 \div 4 = 100\) and \(24 \div 4 = 6\), so the quotient is \(106\). 2. For part b, \(721 = 700 + 21\). Then \(700 \div 7 = 100\) and \(21 \div 7 = 3\), so the quotient is \(103\). 3. For part c, \(636 = 600 + 36\). Then \(600 \div 6 = 100\) and \(36 \div 6 = 6\), so the quotient is \(106\). 4. For part d, \(918 = 900 + 18\). Then \(900 \div 9 = 100\) and \(18 \div 9 = 2\), so the quotient is \(102\).

a) \(106\) b) \(103\) c) \(106\) d) \(102\)

- Compare the divisors within each row. - Use a known quotient to find a related one when the divisor doubles. - In b), compare \(9\) with \(3\). - You may decompose \(360\) into smaller numbers that are easier to divide.

1. a) \(360 \div 2 = 180\). Doubling the divisor gives \(360 \div 4 = 90\), and doubling it again gives \(360 \div 8 = 45\). 2. b) \(360 \div 3 = 120\). Doubling the divisor gives \(360 \div 6 = 60\). Since \(9\) is three times \(3\), \(360 \div 9 = 120 \div 3 = 40\).

a) \(180\), \(90\), \(45\) b) \(120\), \(60\), \(40\)

- Break \(840\) into parts that are easy to divide. - Think about how dividing by \(4\) is related to dividing by \(2\). - Think about how dividing by \(6\) is related to dividing by \(3\). - For part c, use multiplication facts to check each quotient.

1. \(840\div2=420\), so \(840\div4=210\), which is half of \(420\). 2. \(840\div3=280\), so \(840\div6=140\), which is half of \(280\). 3. \(840\div7=120\), and \(840\div8=105\).

a) \(420\) and \(210\) b) \(280\) and \(140\) c) \(120\) and \(105\)

- Find both quotient values before making a comparison. - Use related multiplication facts to help with each division. - Equal quotients need the equality sign, even when the dividends differ.

1. \(240\div4=60\) and \(180\div3=60\), so the quotients are equal. 2. \(560\div7=80\) and \(540\div6=90\), so \(80<90\). 3. \(420\div6=70\) and \(350\div5=70\), so the quotients are equal. 4. \(320\div8=40\) and \(270\div9=30\), so \(40>30\).

1) \(60=60\); \(=\). 2) \(80<90\); \(<\). 3) \(70=70\); \(=\). 4) \(40>30\); \(>\).

- Break each dividend into two numbers that are easy to divide. - Divide each part, then add the partial quotients. - Compare the four results.

1. \(420 \div 3 = (300 \div 3) + (120 \div 3) = 100 + 40 = 140\). 2. \(560 \div 4 = (400 \div 4) + (160 \div 4) = 100 + 40 = 140\). 3. \(700 \div 5 = (500 \div 5) + (200 \div 5) = 100 + 40 = 140\). 4. \(840 \div 6 = (600 \div 6) + (240 \div 6) = 100 + 40 = 140\). 5. All four quotients are equal.

a) \(140\) b) \(140\) c) \(140\) d) \(140\) All four quotients are equal.

- Break each dividend into convenient parts. - Record every quotient before comparing them. - Check a quotient by multiplying it by the divisor.

1. \(480 \div 3 = (300 \div 3) + (180 \div 3) = 100 + 60 = 160\). 2. \(640 \div 4 = (400 \div 4) + (240 \div 4) = 100 + 60 = 160\). 3. \(800 \div 5 = (500 \div 5) + (300 \div 5) = 100 + 60 = 160\). 4. \(780 \div 6 = (600 \div 6) + (180 \div 6) = 100 + 30 = 130\). 5. Part d) is the only quotient that is not \(160\).

d) does not match. \(780 \div 6 = 130\), while the other three quotients are \(160\).

- Compare the divisors \(4\) and \(8\). - Consider what happens to a quotient when the divisor doubles. - Compare the two quotients for one dividend before checking the rest.

1. Dividing by \(4\) gives \(40\), \(80\), \(120\), \(160\), and \(200\). 2. Dividing by \(8\) gives \(20\), \(40\), \(60\), \(80\), and \(100\). 3. Since \(8\) is twice \(4\), dividing the same number by \(8\) gives half the quotient obtained by dividing it by \(4\).

Quotients with \(4\): \(40\), \(80\), \(120\), \(160\), \(200\) Quotients with \(8\): \(20\), \(40\), \(60\), \(80\), \(100\) Each quotient with \(8\) is half the matching quotient with \(4\).

- Break a number into parts that are easy to divide. - Use related multiplication facts. - After dividing by \(2\) and by \(3\), look for a relationship that can help you divide by \(6\).

1. For \(120\): \(120 \div 2 = 60\), \(120 \div 3 = 40\), and \(120 \div 6 = 20\). 2. For \(300\): \(300 \div 2 = 150\), \(300 \div 3 = 100\), and \(300 \div 6 = 50\). 3. For \(420\): \(420 \div 2 = 210\), \(420 \div 3 = 140\), and \(420 \div 6 = 70\). 4. For \(600\): \(600 \div 2 = 300\), \(600 \div 3 = 200\), and \(600 \div 6 = 100\).

The completed table is: <table> <tr><td><b>Starting number</b></td><td>\(120\)</td><td>\(300\)</td><td>\(420\)</td><td>\(600\)</td></tr> <tr><td><b>Divided by \(2\)</b></td><td>\(60\)</td><td>\(150\)</td><td>\(210\)</td><td>\(300\)</td></tr> <tr><td><b>Divided by \(3\)</b></td><td>\(40\)</td><td>\(100\)</td><td>\(140\)</td><td>\(200\)</td></tr> <tr><td><b>Divided by \(6\)</b></td><td>\(20\)</td><td>\(50\)</td><td>\(70\)</td><td>\(100\)</td></tr> </table>

- Use related multiplication facts to divide by \(5\). - To divide by \(2\), think about half of the number. - Multiply each quotient by its divisor to check the table.

1. \(200\div5=40\) and \(200\div2=100\). 2. \(350\div5=70\) and \(350\div2=175\). 3. \(600\div5=120\) and \(600\div2=300\). 4. \(850\div5=170\) and \(850\div2=425\).

<table> <tr><th>Number</th><th>Divided by \(5\)</th><th>Divided by \(2\)</th></tr> <tr><td>\(200\)</td><td>\(40\)</td><td>\(100\)</td></tr> <tr><td>\(350\)</td><td>\(70\)</td><td>\(175\)</td></tr> <tr><td>\(600\)</td><td>\(120\)</td><td>\(300\)</td></tr> <tr><td>\(850\)</td><td>\(170\)</td><td>\(425\)</td></tr> </table>

- Identify the related basic division fact. - Think of the dividend as a number of tens. - Use a multiplication fact to check each quotient.

1. Since \(32 \div 4 = 8\), \(320 \div 4 = 80\). 2. Since \(45 \div 9 = 5\), \(450 \div 9 = 50\). 3. Since \(63 \div 7 = 9\), \(630 \div 7 = 90\). 4. Since \(21 \div 3 = 7\), \(210 \div 3 = 70\). 5. Since \(81 \div 9 = 9\), \(810 \div 9 = 90\).

a) \(80\) b) \(50\) c) \(90\) d) \(70\) e) \(90\)

- Break each dividend into familiar multiples of the divisor. - Make sure the parts add to the original dividend. - Divide each part and add the partial quotients.

1. For part a, \(600 \div 3 = 200\), \(30 \div 3 = 10\), and \(9 \div 3 = 3\). Then \(200 + 10 + 3 = 213\). 2. For part b, \(800 \div 4 = 200\), \(40 \div 4 = 10\), and \(16 \div 4 = 4\). Then \(200 + 10 + 4 = 214\). 3. For part c, \(500 \div 5 = 100\), \(250 \div 5 = 50\), and \(15 \div 5 = 3\). Then \(100 + 50 + 3 = 153\). 4. For part d, \(800 \div 8 = 100\), \(160 \div 8 = 20\), and \(24 \div 8 = 3\). Then \(100 + 20 + 3 = 123\).

a) \(213\) b) \(214\) c) \(153\) d) \(123\)

- Find a large familiar multiple of the divisor. - Divide the remaining part by the same divisor. - Add the partial quotients.

1. For part a, \(520 = 400 + 120\). Then \(400 \div 4 = 100\) and \(120 \div 4 = 30\), so the quotient is \(130\). 2. For part b, \(720 = 600 + 120\). Then \(600 \div 6 = 100\) and \(120 \div 6 = 20\), so the quotient is \(120\). 3. For part c, \(850 = 500 + 350\). Then \(500 \div 5 = 100\) and \(350 \div 5 = 70\), so the quotient is \(170\). 4. For part d, \(910 = 700 + 210\). Then \(700 \div 7 = 100\) and \(210 \div 7 = 30\), so the quotient is \(130\).

a) \(130\) b) \(120\) c) \(170\) d) \(130\)

- Choose a large familiar multiple of the divisor below each dividend. - Divide that part and the remaining part. - Add the partial quotients.

1. For part a, \(100 \div 5 = 20\) and \(35 \div 5 = 7\). Then \(20 + 7 = 27\). 2. For part b, \(120 \div 6 = 20\) and \(48 \div 6 = 8\). Then \(20 + 8 = 28\). 3. For part c, \(240 \div 4 = 60\) and \(12 \div 4 = 3\). Then \(60 + 3 = 63\). 4. For part d, \(320 \div 8 = 40\) and \(24 \div 8 = 3\). Then \(40 + 3 = 43\). 5. For part e, \(420 \div 7 = 60\) and \(35 \div 7 = 5\). Then \(60 + 5 = 65\).

a) \(27\) b) \(28\) c) \(63\) d) \(43\) e) \(65\)

- Break \(840\) into two numbers that are easy to divide by \(4\). - Divide each part by \(4\). - Add the partial quotients.

1. Divide the total number of stickers by the number of friends: \(840 \div 4\). 2. Break \(840\) into compatible numbers: \(840 = 800 + 40\). 3. Divide each part: \(800 \div 4 = 200\) and \(40 \div 4 = 10\). 4. Add the partial quotients: \(200 + 10 = 210\).

Each friend receives \(210\) stickers.

- Work in cents to divide a whole number. - Split the total number of cents into two numbers that are easy to divide by \(3\). - Change your result back into dollars and cents.

1. Change the total price to cents: \(\$7.50=750\) cents. 2. Divide the cents equally among the three children: \(750\div3\). 3. Split the dividend: \(600\div3=200\) and \(150\div3=50\). Together, each child pays \(200+50=250\) cents. 4. \(250\) cents is \(\$2.50\).

\(\$2.50\) per child.

- Break \(780\) into two multiples of \(6\). - Divide each part by \(6\). - Add the partial quotients.

1. Divide the total number of rolls by the number of baskets: \(780 \div 6\). 2. Break \(780\) into compatible numbers: \(780 = 600 + 180\). 3. Divide each part: \(600 \div 6 = 100\) and \(180 \div 6 = 30\). 4. Add the partial quotients: \(100 + 30 = 130\).

Each basket contains \(130\) rolls.

- Count the dots across a complete row. - Count the complete rows separately from the short row. - Check that the leftover dots cannot make another full row.

1. The array has \(35\) dots and \(6\) columns. Five complete rows contain \(5\times6=30\) dots. 2. There are \(35-30=5\) dots in the incomplete row. 3. Thus \(35\div6=5\) remainder \(5\).

\(5\) full rows, \(5\) dots left over; \(35\div6=5\) remainder \(5\)

- Count how many dots make one full row. - Count the complete rows before considering the final row. - Does the short row contain enough students for another team?

1. The array shows \(42\) students in rows of \(5\). 2. Eight complete rows contain \(8\times5=40\) students. 3. The final row has \(42-40=2\) students, so \(42\div5=8\) remainder \(2\).

\(8\) complete teams and \(2\) students left over

- Count how many medals make one row and thus one team prize. - How many full rows are in the array? - Check by multiplying the number of teams by the medals per team.

1. The diagram has \(42\) dots arranged in rows of \(3\). 2. There are \(14\) complete rows, so \(42\div3=14\). 3. Check: \(14\times3=42\).

\(14\) teams; \(42\div3=14\)

- Count the dots in one complete row and the number of full rows. - Count the dots in the incomplete row. - Your check should account for every dot in the array.

1. The array shows \(77\) tickets, with \(9\) tickets per full row. 2. There are \(8\) full rows: \(8\times9=72\). 3. The short row contains \(5\) tickets, so \(77\div9=8\) remainder \(5\). 4. Check: \(8\times9+5=77\).

\(8\) full envelopes and \(5\) tickets left over Check: \(8\times9+5=77\)

- Find how many complete equal packets can be made without using more clay than is available. - After accounting for the full packets, check whether any clay remains.

1. Find the greatest multiple of \(8\) that does not exceed \(50\): \(6 \times 8 = 48\). 2. The club can make \(6\) full packets. 3. Subtract the packed mass: \(50 - 48 = 2\).

The club can make \(6\) full packets, with \(2\,\text{g}\) of clay left.

- Follow the quotient digits from left to right. - Each subtrahend is the divisor multiplied by the quotient digit for that step. - Match the middle quotient digit to the subtraction under the partial dividend \(6\).

1. The calculation is \(864\div4=216\). 2. After \(8\div4=2\), the next partial dividend is \(6\). 3. The divisor \(4\) fits into \(6\) once, so the quotient digit is \(1\) and the subtrahend is \(1\times4=4\). 4. The final step uses \(24\div4=6\), giving quotient \(216\).

a) \(216\) b) Because the quotient digit in that place is \(1\), so \(1\times4=4\).

- Track the remainder from the first subtraction before bringing down the next digit. - Compare nearby multiples of \(5\) with the partial dividend \(23\). - Continue the same process for the final digit.

1. The calculation is \(735\div5\). 2. After \(7-5=2\), bring down the \(3\) to make \(23\). 3. The divisor \(5\) fits into \(23\) four times because \(4\times5=20\) and \(5\times5=25\) is too large. 4. Continuing with \(35\div5=7\) gives quotient \(147\).

The quotient digit \(4\) comes from \(23\div5\), and the quotient is \(147\).

- Match each quotient digit with the subtraction step directly below its partial dividend. - Each subtrahend is a multiple of the divisor. - Read the multiplication facts in the same left-to-right order as the quotient digits.

1. The calculation is \(936\div3=312\). 2. The first quotient digit \(3\) uses \(3\times3=9\). 3. The next quotient digit \(1\) uses \(1\times3=3\). 4. The final quotient digit \(2\) uses \(2\times3=6\).

The facts are \(3\times3=9\), \(1\times3=3\), and \(2\times3=6\). The quotient is \(312\).

- Read the quotient and remainder from the written work. - Decide whether the leftover guests need another table. - The remainder tells how many guests cannot join a full table.

1. Read the quotient \(236\) and remainder \(3\) in the division work. 2. The quotient gives \(236\) completely full tables. 3. Three more guests need one additional table, making \(237\) tables in all. 4. The last table seats \(3\) guests.

a) \(236\) full tables b) \(237\) tables altogether c) \(3\) guests at the final table

- Read the quotient and remainder from the written calculation. - Decide which one counts cartons and which one counts loose notebooks. - Compare the remainder with the divisor to find the number still needed.

1. Read the quotient \(164\) and remainder \(3\) from the written division. 2. The quotient means \(164\) full cartons can be filled; \(3\) notebooks remain. 3. Each carton holds \(6\) notebooks, so another full carton needs \(6-3=3\) more.

a) \(164\) full cartons b) \(3\) notebooks c) \(3\) additional notebooks

- Read the first subtraction in the displayed calculation. - What digit is brought down next? - Explain the transition before reading the final quotient.

1. The first step is \(13-12=1\), leaving a remainder of \(1\). 2. Bring down the next digit, \(2\), next to the remaining \(1\) ten. The next number to divide is \(12\). 3. The shown quotient is \(221\).

a) \(1\) b) \(12\), formed by bringing down \(2\) next to the remaining \(1\). c) \(221\)

- When you cannot share all the hundreds equally, what can you do with any leftovers? - Remember that one hundred can be regrouped without changing its value. - Check the student's claim by multiplying your quotient by \(3\).

1. Share \(7\) hundreds among \(3\) groups: each gets \(2\) hundreds, with \(1\) hundred left. 2. Trade the leftover hundred for \(10\) tens. Each group gets \(3\) tens, with \(1\) ten left. 3. Trade the leftover ten for \(10\) ones and add the \(8\) ones. Then \(18\div3=6\) ones per group. 4. So \(708\div3=236\). The quotient has \(3\) tens, not \(0\).

No. \(708\div3=236\). Regrouping a leftover hundred creates tens in the quotient.

- Share the hundreds, but keep track of what cannot yet be shared equally. - Trade leftover hundreds for tens before dividing the tens column. - Repeat the idea with any leftover tens before dividing ones.

1. Give each of the \(6\) groups \(1\) hundred. That uses \(600\) of the \(732\), leaving \(132\): \(1\) hundred, \(3\) tens, and \(2\) ones. 2. Trade the remaining hundred for \(10\) tens. Divide the \(13\) tens by \(6\): each group gets \(2\) tens, with \(1\) ten left. 3. Trade the ten for \(10\) ones and include \(2\) more ones. Divide \(12\) ones by \(6\): each gets \(2\) ones. 4. The quotient is \(122\).

\(132\) remains to divide. \(732\div6=122\).

- Use each partial dividend shown in the written work. - The difference after subtracting the middle row becomes part of the next partial dividend. - Check each quotient digit against the divisor.

1. \(8\div4=2\). Subtract \(8\) and bring down the next digit, \(6\). 2. \(6\div4=1\). The middle subtraction entry is \(4\), leaving \(2\) tens. 3. Bring down the \(4\) ones to make \(24\). Divide \(24\div4=6\) and subtract \(24\), leaving remainder \(0\). 4. The completed quotient is \(216\).

Quotient: \(216\). Missing subtraction entries: \(4\) and \(24\). Remainder: \(0\). In the middle step, \(6-4=2\) tens remain. Bring down the \(4\) ones to make \(24\) ones.

- Find the quotient and remainder for each dividend in order. - Notice how the remainder changes when the dividend increases by \(1\). - A remainder cannot be equal to or greater than the divisor.

1. The results are \(42 \div 7 = 6\) remainder \(0\), \(43 \div 7 = 6\) remainder \(1\), \(44 \div 7 = 6\) remainder \(2\), \(45 \div 7 = 6\) remainder \(3\), \(46 \div 7 = 6\) remainder \(4\), \(47 \div 7 = 6\) remainder \(5\), \(48 \div 7 = 6\) remainder \(6\), and \(49 \div 7 = 7\) remainder \(0\). 2. The remainders \(0\) through \(6\) occur. 3. The greatest possible remainder is \(6\), because a remainder must be less than the divisor \(7\).

The remainders are \(0\), \(1\), \(2\), \(3\), \(4\), \(5\), \(6\), and \(0\). The greatest possible remainder is \(6\) because a remainder must be less than the divisor.

- Start with a nearby multiple of \(9\) for each dividend. - Imagine putting the leftovers into another group of \(9\). When would that be possible? - Test several possible leftovers to decide where the list of remainders must stop.

1. \(79=9\times8+7\), \(80=9\times8+8\), \(81=9\times9+0\), and \(82=9\times9+1\). 2. There are \(9\) possible remainders: \(0,1,2,3,4,5,6,7,8\). 3. If \(10\) were left over, another full group of \(9\) could be made, leaving \(1\) instead. The remainder cannot be as large as the divisor.

a) \(8\) remainder \(7\); \(8\) remainder \(8\); \(9\) remainder \(0\); \(9\) remainder \(1\). b) \(9\) remainders: \(0,1,2,3,4,5,6,7,8\). c) Another group of \(9\) can be made from \(10\) leftovers, so \(10\) cannot be the remainder.

- In a) and b), find the number of full groups before counting what remains. - In c), use the remainder from b) to decide what must change. - In d), think about when leftovers would make another full group.

1. \(23=4\times5+3\), so \(3\) notebooks remain in a). 2. \(23=5\times4+3\), so \(3\) remain in b). 3. The next number, \(24\), has one more leftover when divided by \(5\): \(24=5\times4+4\). It is the least number greater than \(23\) that works. 4. The largest possible remainder when dividing by \(5\) is \(4\); with \(5\) leftovers, another full group could be made.

a) \(3\) notebooks. b) \(3\) notebooks. c) \(24\) notebooks. d) \(4\) notebooks.

- Divide the number of students by the number of seats on each bench. - What do the quotient and remainder represent? - Do the students in the remainder need another bench?

1. Divide to find the quotient and remainder: \(38 \div 4 = 9\) remainder \(2\). 2. The quotient shows that \(9\) benches will be completely full. 3. The remainder shows that \(2\) students will sit on one more bench. 4. Add that bench to the \(9\) full benches: \(9 + 1 = 10\).

a) \(9\) benches will be completely full. b) \(2\) students will sit on the last bench. c) \(10\) benches are needed in all.

- First find the number when you know twice that number. - After you find the number, how do you find half of it? - How many times as large is twice a number as half of the same number? - Can you find the target value in one step by dividing by a single number?

To move from twice a number to half of that number, divide by \(2\) and then divide by \(2\) again. This is the same as dividing by \(4\). 1. \(480 \div 2 = 240\), and \(240 \div 2 = 120\). 2. \(760 \div 2 = 380\), and \(380 \div 2 = 190\). 3. \(1200 \div 2 = 600\), and \(600 \div 2 = 300\). 4. \(3400 \div 2 = 1700\), and \(1700 \div 2 = 850\). 5. \(6800 \div 2 = 3400\), and \(3400 \div 2 = 1700\).

Half of each original number is \(120\), \(190\), \(300\), \(850\), and \(1700\).

- Look for the parts \(800\), \(80\), and \(16\) inside the larger dividends. - Divide each part by \(8\), and then add the partial quotients. - Which of the first three dividends can be combined to make \(896\)?

1. Find the first three quotients: \(800 \div 8 = 100\), \(80 \div 8 = 10\), and \(16 \div 8 = 2\). 2. For d), decompose \(880\) as \(800 + 80\): \(100 + 10 = 110\). 3. For e), decompose \(896\) as \(800 + 80 + 16\): \(100 + 10 + 2 = 112\). 4. For f), decompose \(816\) as \(800 + 16\): \(100 + 2 = 102\).

a) \(100\) b) \(10\) c) \(2\) d) \(110\) e) \(112\) f) \(102\)

- Split each dividend into a large multiple of \(100\) and a smaller remainder. - Make sure both parts are divisible by the divisor. - Add the two partial quotients.

1. For a), decompose \(321\) as \(300 + 21\). Then \(300 \div 3 = 100\) and \(21 \div 3 = 7\), so the quotient is \(107\). 2. For b), decompose \(642\) as \(600 + 42\). Then \(600 \div 6 = 100\) and \(42 \div 6 = 7\), so the quotient is \(107\). 3. For c), decompose \(436\) as \(400 + 36\). Then \(400 \div 4 = 100\) and \(36 \div 4 = 9\), so the quotient is \(109\).

a) \(107\) b) \(107\) c) \(109\)

- Each division has been broken into easier partial divisions. - Add the partial dividends to find the missing dividend. - For b), find each partial quotient before adding them.

1. For a), add the partial dividends: \(500 + 45 = 545\). Therefore, \(545 \div 5 = 109\). 2. For b), add the partial dividends: \(700 + 70 + 7 = 777\). 3. Find the partial quotients: \(700 \div 7 = 100\), \(70 \div 7 = 10\), and \(7 \div 7 = 1\). 4. Add them: \(100 + 10 + 1 = 111\). Therefore, \(777 \div 7 = 111\).

a) \(545 \div 5 = 109\) b) \(777 \div 7 = 100 + 10 + 1 = 111\)

- Multiply each given quotient by its divisor. - Compare every check product with the dividend. - For any incorrect equation, find a quotient that passes the same check.

1. In a), \(208\times3=624\), so the equation is correct. 2. In b), \(149\times5=745\), so the equation is correct. 3. In c), \(14\times9=126\), not \(936\), so the equation is incorrect. 4. The correct quotient is \(104\): \(104\times9=936\).

a) Correct; \(208\times3=624\). b) Correct; \(149\times5=745\). c) Incorrect; \(14\times9=126\neq936\). The correct quotient is \(104\), since \(104\times9=936\).

- Divide the total mass equally among the \(9\) plums. - Find the greatest multiple of \(9\) that does not exceed \(400\). - Use the quotient, not a separate rounding convention, for the requested estimate.

1. Divide the total mass by the number of plums: \(400\div9\). 2. \(400\div9=44\) remainder \(4\). 3. Using the whole-number quotient as directed, one plum has a mass of about \(44\,\text{g}\).

About \(44\,\text{g}\).

- Break \(456\) into two numbers that are both multiples of \(4\). - A multiple of \(100\) may be a useful first part. - Add the partial quotients at the end.

1. Decompose \(456\) as \(400 + 56\), because both parts are divisible by \(4\). 2. Divide the first part: \(400 \div 4 = 100\). 3. Divide the second part: \(56 \div 4 = 14\). 4. Add the partial quotients: \(100 + 14 = 114\).

\(114\)

- Multiply the requested quotient by possible one-digit divisors. - Check whether the product is inside the given range. - Make sure each divisor is different.

1. Search for one-digit divisors whose products with \(15\) lie between \(100\) and \(200\). 2. The divisors \(7\), \(8\), and \(9\) give \(7\times15=105\), \(8\times15=120\), and \(9\times15=135\). 3. Reverse these facts to get \(105\div7=15\), \(120\div8=15\), and \(135\div9=15\).

\(105\div7=15\), \(120\div8=15\), and \(135\div9=15\)

- Choose divisors from \(3\) through \(9\). - Multiply each divisor by \(42\) to find a dividend. - Check each division equation.

1. Choose divisors such as \(3\), \(4\), and \(5\). 2. Find the dividends: \(3 \times 42 = 126\), \(4 \times 42 = 168\), and \(5 \times 42 = 210\). 3. The equations are \(126 \div 3 = 42\), \(168 \div 4 = 42\), and \(210 \div 5 = 42\).

One possible set is \(126 \div 3 = 42\), \(168 \div 4 = 42\), and \(210 \div 5 = 42\).

- First decide whether a related basic fact makes the quotient easy to find mentally. - Try breaking the dividend into hundreds, tens, and ones. - Use the standard algorithm when the partial divisions are difficult to organize mentally.

1. For a), use mental math: \(56 \div 7 = 8\), so \(560 \div 7 = 80\). 2. For b), use partial quotients: \(800 \div 4 = 200\), \(40 \div 4 = 10\), and \(4 \div 4 = 1\). Then \(200 + 10 + 1 = 211\). 3. For c), use the standard division algorithm. Since \(1<6\), begin with \(13\). Divide \(13 \div 6 = 2\text{ R }1\). Bring down the next \(4\): \(14 \div 6 = 2\text{ R }2\). Bring down the last \(4\): \(24 \div 6 = 4\). Therefore, \(1344 \div 6 = 224\).

a) \(80\) b) \(211\) c) \(224\)

- A nonzero remainder means an equal sharing leaves some objects unassigned. - Use the standard division algorithm one place at a time. - Check that each remainder is less than the divisor.

1. For a), \(8008 \div 8=1001\). 2. For b), \(8008 \div 3=2669\text{ R }1\). Because the remainder is \(1\), an equal distribution with none left over is not possible. 3. For c), \(8008 \div 6=1334\text{ R }4\), so the remainder is \(4\).

a) \(1001\) b) No. \(8008 \div 3=2669\text{ R }1\), so one marble is left over. c) The remainder is \(4\).

- Divide each number by \(2\). - Break a number into parts that are easy to halve if needed. - Compare the thousands and hundreds places when ordering.

1. Find the halves: \(7400 \div 2 = 3700\), \(3200 \div 2 = 1600\), \(9600 \div 2 = 4800\), and \(5800 \div 2 = 2900\). 2. Order the results: \(1600 < 2900 < 3700 < 4800\).

The halves are \(3700\), \(1600\), \(4800\), and \(2900\), in the order of the given numbers. From least to greatest: \(1600\), \(2900\), \(3700\), \(4800\).

- Look for earlier dividends that can be combined to make a later dividend. - Divide convenient parts separately and add the partial quotients. - Reuse earlier results to solve the final division efficiently.

1. \(6000 \div 6 = 1000\). 2. \(420 \div 6 = 70\). 3. Since \(6420 = 6000 + 420\), \(6420 \div 6 = 1000 + 70 = 1070\). 4. \(18 \div 6 = 3\). 5. Since \(6438 = 6420 + 18\), \(6438 \div 6 = 1070 + 3 = 1073\).

a) \(1000\) b) \(70\) c) \(1070\) d) \(3\) e) \(1073\)

- Choose a one-digit divisor first. - Use multiplication as the inverse of division. - Multiply the required quotient by your chosen divisor to find the dividend.

1. Choose four different one-digit divisors, such as \(2,3,4,5\). 2. Multiply \(1200\) by each divisor to find the corresponding dividend: \(1200\times 2=2400\), \(1200\times 3=3600\), \(1200\times 4=4800\), and \(1200\times 5=6000\). 3. Therefore, four valid equations are \(2400\div 2=1200\), \(3600\div 3=1200\), \(4800\div 4=1200\), and \(6000\div 5=1200\).

One possible answer is: \(2400\div 2=1200\) \(3600\div 3=1200\) \(4800\div 4=1200\) \(6000\div 5=1200\)

- Notice how each large dividend is built from easier parts. - Divide the parts separately, and then add the partial quotients. - How do the first two equations in each part combine to form the final equation?

1. For a), \(2000 \div 2 = 1000\) and \(18 \div 2 = 9\). Add the partial quotients: \(1000 + 9 = 1009\). 2. For b), combine the partial dividends: \(9000 + 630 = 9630\). Therefore, \(9630 \div 9 = 1070\). 3. For c), add the partial quotients: \(100 + 9 = 109\). Therefore, \(545 \div 5 = 109\).

a) \(2000 \div 2 = 1000\), \(18 \div 2 = 9\), and \(2018 \div 2 = 1009\) b) \(9630 \div 9 = 1070\) c) \(545 \div 5 = 109\)

- Look for a large part related to a basic multiplication fact and place value. - Make sure the remaining part is also divisible by the divisor. - Add the two partial quotients.

1. For a), decompose \(4212\) as \(4200 + 12\). Then \(4200 \div 6 = 700\) and \(12 \div 6 = 2\), so the quotient is \(702\). 2. For b), decompose \(6456\) as \(6400 + 56\). Then \(6400 \div 8 = 800\) and \(56 \div 8 = 7\), so the quotient is \(807\). 3. For c), decompose \(2135\) as \(2100 + 35\). Then \(2100 \div 7 = 300\) and \(35 \div 7 = 5\), so the quotient is \(305\).

a) \(702\) b) \(807\) c) \(305\)

- Which proposed parts are easy multiples of \(6\)? - Use related basic multiplication facts to identify convenient dividends. - For c), look for a nearby multiple of \(5\) that is easy to divide.

1. For a), \(3600 + 18\) is more efficient because both parts are immediately recognizable multiples of \(6\). 2. For b), \(3600 \div 6 = 600\) and \(18 \div 6 = 3\). Therefore, \(3618 \div 6 = 603\). 3. For c), decompose \(4540\) as \(4500 + 40\). Then \(4500 \div 5 = 900\) and \(40 \div 5 = 8\), so the quotient is \(908\).

a) \(3600 + 18\), because both parts are easy to divide by \(6\) b) \(603\) c) \(908\)

- Which dividend is easiest to divide by \(6\)? - How much greater or less are the other dividends? - How does the quotient change when the dividend changes by \(6\)?

1. The easiest problem is \(1800 \div 6 = 300\). 2. The dividend \(1842\) is \(42\) greater than \(1800\). Since \(42 \div 6 = 7\), add \(7\): \(300 + 7 = 307\). 3. The dividend \(1794\) is \(6\) less than \(1800\). Since \(6 \div 6 = 1\), subtract \(1\): \(300 - 1 = 299\).

a) \(307\) b) \(300\) c) \(299\)

- Find the missing part that combines with the given part to make the original dividend. - Divide both parts separately. - Add the partial quotients.

1. For a), decompose \(2832\) as \(2800 + 32\). Then \(2800 \div 4 = 700\) and \(32 \div 4 = 8\), so the quotient is \(708\). 2. For b), decompose \(8127\) as \(8100 + 27\). Then \(8100 \div 9 = 900\) and \(27 \div 9 = 3\), so the quotient is \(903\).

a) \(2832 \div 4 = (2800 \div 4) + (32 \div 4) = 700 + 8 = 708\) b) \(8127 \div 9 = (8100 \div 9) + (27 \div 9) = 900 + 3 = 903\)

- Look for parts of each dividend that divide evenly by the divisor. - Pay attention to place values where the quotient may need a zero. - Use the standard algorithm when a convenient decomposition is not easy to see.

1. For a), use mental math with the basic fact \(45 \div 9 = 5\) and place value: \(4500 \div 9 = 500\). 2. For b), use partial quotients: \(4599 \div 9 = 4500 \div 9 + 99 \div 9 = 500 + 11 = 511\). 3. For c), use the standard algorithm. Divide \(42 \div 6 = 7\). Bring down \(1\); since \(1<6\), write \(0\) in the tens place and carry the remainder \(1\). Bring down \(8\) to make \(18\), and \(18 \div 6 = 3\). Therefore, \(4218 \div 6 = 703\). 4. For d), use partial quotients: \(4832 \div 8 = 4800 \div 8 + 32 \div 8 = 600 + 4 = 604\).

a) \(500\) b) \(511\) c) \(703\) d) \(604\)

- Divide the total number of apples by the number in each bag. - The quotient represents full bags. - The remainder represents apples left over.

1. Divide: \(4589 \div 7\). 2. The standard division algorithm gives \(655\text{ R }4\). 3. Therefore, \(655\) full bags can be packed, with \(4\) apples left. Check: \(655 \times 7+4=4589\).

\(655\) full bags can be packed, and \(4\) apples will be left over.

- First divide the total award by the number of schools. - Use that quotient as the dividend in the second division. - Check each quotient by multiplication.

1. Divide the award among four schools: \(9984\div4=2496\). 2. Divide one school's share among eight classes: \(2496\div8=312\). 3. Each class receives \(\$312\).

Each class receives \(\$312\).

- Estimate each quotient using a nearby compatible number. - Compare the number of digits in each estimate with the stated quotient. - Look for the result that differs greatly from its estimate.

1. For a), \(3456 \div 9\approx3600 \div 9=400\). The stated quotient \(384\) is reasonable. 2. For b), \(1944 \div 6\approx1800 \div 6=300\). The stated quotient \(324\) is reasonable. 3. For c), \(5565 \div 7\approx5600 \div 7=800\). The stated quotient \(95\) is far too small. The exact quotient is \(795\).

Equation c) is incorrect. Since \(5565 \div 7\approx800\), the quotient must be in the hundreds, not \(95\).

- Choose a nearby number that is easy to divide by the divisor. - Then use the standard division algorithm for the exact quotient. - Compare each exact result with its estimate.

1. For a), use the compatible number \(1680\): \(1638 \div 7\approx1680 \div 7=240\). For the exact quotient, begin with \(16\). Then \(16 \div 7=2\) with remainder \(2\). Bring down the \(3\) to make \(23\), and \(23 \div 7=3\) with remainder \(2\). Bring down the \(8\) to make \(28\), and \(28 \div 7=4\). Therefore, \(1638 \div 7=234\). 2. For b), use the compatible number \(2250\): \(2304 \div 9\approx2250 \div 9=250\). For the exact quotient, begin with \(23\). Then \(23 \div 9=2\) with remainder \(5\). Bring down the \(0\) to make \(50\), and \(50 \div 9=5\) with remainder \(5\). Bring down the \(4\) to make \(54\), and \(54 \div 9=6\). Therefore, \(2304 \div 9=256\).

a) Estimate: \(240\); exact quotient: \(234\) b) Estimate: \(250\); exact quotient: \(256\)

- Align each division carefully in the standard algorithm. - Compare consecutive dividends. - Divide the change in the dividend by \(4\). - Look for that change in the quotients.

1. Part a: \(24\div 4=6\), so write \(6\) in the hundreds place. Since \(3<4\), write \(0\) in the tens place. Bring down \(6\) to make \(36\), and \(36\div 4=9\). Thus, \(2436\div 4=609\). 2. Part b: \(28\div 4=7\). Since \(3<4\), write \(0\), then use \(36\div 4=9\). Thus, \(2836\div 4=709\). 3. Part c: \(32\div 4=8\). Since \(3<4\), write \(0\), then use \(36\div 4=9\). Thus, \(3236\div 4=809\). 4. Part d: \(36\div 4=9\). Since \(3<4\), write \(0\), then use \(36\div 4=9\). Thus, \(3636\div 4=909\). 5. The dividend increases by \(400\) each time. Since \(400\div 4=100\), the quotient increases by \(100\) each time.

a) \(609\) b) \(709\) c) \(809\) d) \(909\) Each increase of \(400\) in the dividend increases the quotient by \(100\).

- Use the standard division algorithm for each quotient. - Compare the divisors from one part to the next. - Compare each quotient with the previous quotient. - Use multiplication to check the last quotient.

1. Part a: \(5\div 2=2\) remainder \(1\); \(11\div 2=5\) remainder \(1\); \(18\div 2=9\); and \(4\div 2=2\). Therefore, \(5184\div 2=2592\). 2. Part b: \(5\div 4=1\) remainder \(1\); \(11\div 4=2\) remainder \(3\); \(38\div 4=9\) remainder \(2\); and \(24\div 4=6\). Therefore, \(5184\div 4=1296\). 3. Part c: Since \(5<8\), use \(51\). Then \(51\div 8=6\) remainder \(3\); \(38\div 8=4\) remainder \(6\); and \(64\div 8=8\). Therefore, \(5184\div 8=648\). 4. Check part c: \(648\times 8=5184\). 5. The dividend stays fixed while the divisor doubles each time, so the quotient is halved each time.

a) \(2592\) b) \(1296\) c) \(648\) Check: \(648\times 8=5184\). When the divisor doubles and the dividend stays fixed, the quotient is halved.

- Choose a nearby compatible number divisible by \(6\). - Watch for zeros that must be written in the quotient. - Compare each exact quotient with its estimate.

1. For a), \(66 \div 6\approx60 \div 6=10\). The exact quotient is \(11\). 2. For b), \(606 \div 6\approx600 \div 6=100\). The exact quotient is \(101\). 3. For c), \(6006 \div 6\approx6000 \div 6=1000\). The exact quotient is \(1001\).

a) Estimate: \(10\); exact quotient: \(11\) b) Estimate: \(100\); exact quotient: \(101\) c) Estimate: \(1000\); exact quotient: \(1001\)

- Complete both divisions through the final remainder. - Compare only the remainders. - A remainder must always be less than its divisor.

1. For a), begin with \(14\) because \(1<5\). Then \(14 \div 5=2\) with remainder \(4\). Bring down the \(2\) to make \(42\). Then \(42 \div 5=8\) with remainder \(2\). Bring down the \(7\) to make \(27\). Then \(27 \div 5=5\) with remainder \(2\). Therefore, \(1427 \div 5=285\text{ R }2\). 2. For b), \(9 \div 9=1\). Bring down the \(8\). Since \(9\) goes into \(8\) zero times, write \(0\) in the hundreds place of the quotient. Bring down the \(3\) to make \(83\). Then \(83 \div 9=9\) with remainder \(2\). Bring down the \(5\) to make \(25\). Then \(25 \div 9=2\) with remainder \(7\). Therefore, \(9835 \div 9=1092\text{ R }7\). 3. Since \(7>2\), part b) has the greater remainder.

a) \(285\text{ R }2\) b) \(1092\text{ R }7\) Part b) has the greater remainder.

- Find a nearby multiple of \(7\) for each number. - Choose the greatest multiple that does not exceed the number you are dividing. - Check that any remainder is smaller than \(7\).

1. \(560\div7=80\). 2. \(566=7\times80+6\), so \(566\div7\) gives \(80\) remainder \(6\). 3. \(558=7\times79+5\), so \(558\div7\) gives \(79\) remainder \(5\).

a) \(80\). b) \(80\) remainder \(6\). c) \(79\) remainder \(5\).

- Calculate the divisions in order within each part. - Track the quotient and remainder separately. - Remember that a remainder must be less than the divisor. - Notice what happens when another item completes a full group.

1. Part a gives \(70\), \(70\text{ R }1\), and \(70\text{ R }2\). 2. Part b gives \(52\text{ R }2\), \(52\text{ R }3\), and \(53\). 3. Part c gives \(42\), \(42\text{ R }1\), and \(42\text{ R }2\). 4. As the dividend increases by \(1\), the remainder increases by \(1\). When the remainder would equal the divisor, it becomes one more whole group, so the quotient increases by \(1\) and the remainder returns to \(0\).

a) \(70,70\text{ R }1,70\text{ R }2\) b) \(52\text{ R }2,52\text{ R }3,53\) c) \(42,42\text{ R }1,42\text{ R }2\) Increasing the dividend by \(1\) usually increases the remainder by \(1\). When the remainder reaches the divisor, the quotient increases by \(1\) and the remainder returns to \(0\).

- Divide the total number of tulips by the number in each bouquet. - Interpret the quotient and remainder in the context. - Compare the remainder with the number needed for one full bouquet.

1. Divide: \(345 \div 7=49\text{ R }2\). 2. The quotient shows that \(49\) complete bouquets can be made. 3. The remainder shows that \(2\) tulips are left over. 4. One bouquet needs \(7\) tulips, so \(7-2=5\) more tulips are needed.

a) \(49\) complete bouquets b) \(2\) tulips left over c) \(5\) more tulips

- Reverse the division using multiplication. - For a division with remainder, use divisor times quotient plus remainder. - Then perform the division in part c).

1. For a), use \(\text{dividend}=\text{divisor} \times \text{quotient}+\text{remainder}\): \(6 \times 25+3=153\). 2. For b), \(8 \times 15=120\). 3. For c), \(120 \div 7=17\text{ R }1\), so the remainder is \(1\).

a) \(153\) b) \(120\) c) The remainder is \(1\).

- Work through the divisions one at a time. - Check each result by multiplying the quotient by the divisor and adding the remainder.

1. \(3361 \div 2 = 1680\) remainder \(1\). 2. \(3361 \div 3 = 1120\) remainder \(1\). 3. \(3361 \div 5 = 672\) remainder \(1\). 4. \(3361 \div 7 = 480\) remainder \(1\). 5. Every division has a remainder of \(1\).

a) \(3361 \div 2 = 1680\) remainder \(1\); \(3361 \div 3 = 1120\) remainder \(1\); \(3361 \div 5 = 672\) remainder \(1\); \(3361 \div 7 = 480\) remainder \(1\). b) Every division has a remainder of \(1\).

- Check whether the dividend can be split into friendly parts that both divide evenly. - A written algorithm may be clearer when remainders must be carried through several place values.

1. For a), mental math is efficient. Decompose \(840\) as \(800 + 40\): \(840 \div 4 = 800 \div 4 + 40 \div 4 = 200 + 10 = 210\). 2. For b), use the standard algorithm. Divide \(9 \div 7 = 1\text{ R }2\). Bring down \(5\) to make \(25\), and divide \(25 \div 7 = 3\text{ R }4\). Bring down \(2\) to make \(42\), and divide \(42 \div 7 = 6\). Therefore, \(952 \div 7 = 136\).

a) \(210\) b) \(136\)

- Find all six quotients and remainders. - Compare the remainders after solving. - Check each result with \((\text{quotient} \times \text{divisor}) + \text{remainder} = \text{dividend}\).

1. The results are: a) \(2\) remainder \(2\); b) \(9\) remainder \(2\); c) \(8\) remainder \(2\); d) \(9\) remainder \(2\); e) \(9\) remainder \(3\); f) \(9\) remainder \(2\). 2. Parts a, b, c, d, and f have remainder \(2\).

a) \(2\) remainder \(2\) b) \(9\) remainder \(2\) c) \(8\) remainder \(2\) d) \(9\) remainder \(2\) e) \(9\) remainder \(3\) f) \(9\) remainder \(2\) Parts a), b), c), d), and f) have remainder \(2\).

- Solve each division problem and record its remainder. - Compare the remainders after you have solved all five problems. - Look for the one remainder that is different from the others.

1. The results are: a) \(5\) remainder \(2\); b) \(6\) remainder \(2\); c) \(7\) remainder \(2\); d) \(6\) remainder \(5\); e) \(7\) remainder \(2\). 2. Parts a, b, c, and e all have remainder \(2\).

a) \(5\) remainder \(2\) b) \(6\) remainder \(2\) c) \(7\) remainder \(2\) d) \(6\) remainder \(5\) e) \(7\) remainder \(2\) Parts a), b), c), and e) have the same remainder, \(2\).

- For part a, find the greatest multiple of \(7\) that is less than \(44\). - To work backward from a quotient and remainder, multiply the quotient by the divisor. - Then add the remainder.

1. For part a, \(6 \times 7 = 42\), and \(44 - 42 = 2\). Therefore, \(44 \div 7 = 6\) remainder \(2\). 2. For part b, multiply the quotient by the divisor and add the remainder: \(6 \times 5 + 2 = 30 + 2 = 32\).

a) \(44 \div 7 = 6\) remainder \(2\) b) \(32 \div 5 = 6\) remainder \(2\)

- Find the quotient and remainder for each problem. - Write each remainder beside its problem. - Match the problems that have equal remainders.

1. For part a, \(38 = 4 \times 9 + 2\), so the remainder is \(2\). 2. For part b, \(26 = 6 \times 4 + 2\), so the remainder is \(2\). 3. For part c, \(50 = 7 \times 7 + 1\), so the remainder is \(1\). 4. For part d, \(19 = 6 \times 3 + 1\), so the remainder is \(1\). 5. Parts a and b have remainder \(2\). Parts c and d have remainder \(1\).

a) \(4\) remainder \(2\) b) \(6\) remainder \(2\) c) \(7\) remainder \(1\) d) \(6\) remainder \(1\) Parts a) and b) have the same remainder. Parts c) and d) also have the same remainder.

- Evaluate both quotients in each comparison. - Record the intermediate results. - For the same dividend, compare how the divisor affects the quotient.

1. a) \(84 \div 4 = 21\) and \(66 \div 3 = 22\), so \(21 < 22\). 2. b) \(72 \div 3 = 24\) and \(96 \div 4 = 24\), so the quotients are equal. 3. c) \(95 \div 5 = 19\) and \(68 \div 4 = 17\), so \(19 > 17\). 4. d) \(100 \div 2 = 50\) and \(100 \div 4 = 25\), so \(50 > 25\).

a) \(<\) b) \(=\) c) \(>\) d) \(>\)

- Solve both division problems separately. - Break each dividend into familiar multiples of the divisor. - Compare the two quotients.

1. Break \(132\) into \(120 + 12\). Then \(120 \div 6 = 20\) and \(12 \div 6 = 2\), so \(132 \div 6 = 22\). 2. Break \(176\) into \(160 + 16\). Then \(160 \div 8 = 20\) and \(16 \div 8 = 2\), so \(176 \div 8 = 22\). 3. The quotients are equal.

\(132 \div 6 = 22\) and \(176 \div 8 = 22\). The quotients are equal.

- Find a large multiple of \(8\) close to \(584\). - Divide that part and the remaining part by \(8\). - Add the partial quotients.

1. Write the division problem \(584 \div 8\). 2. Break \(584\) into \(560 + 24\). 3. Divide each part: \(560 \div 8 = 70\) and \(24 \div 8 = 3\). 4. Add the partial quotients: \(70 + 3 = 73\).

There are \(73\) groups of \(8\) in \(584\).

- For part a, break \(168\) into numbers that are easy to divide by \(6\). - For part b, find a different way to break apart \(168\) so both parts are easy to divide by \(8\). - When the same number of bulbs is divided among more boxes, should each box get more or fewer bulbs?

1. For part a, break \(168\) into compatible numbers: \(168 = 120 + 48\). 2. Divide each part by \(6\): \(120 \div 6 = 20\) and \(48 \div 6 = 8\). Then add: \(20 + 8 = 28\). 3. For part b, use a different decomposition: \(168 = 160 + 8\). 4. Divide each part by \(8\): \(160 \div 8 = 20\) and \(8 \div 8 = 1\). Then add: \(20 + 1 = 21\).

a) Each of the \(6\) boxes gets \(28\) tulip bulbs. b) Each of the \(8\) boxes would get \(21\) tulip bulbs.

- How many children are there altogether? - Determine how many complete groups of nine can be made. - What does the remainder represent in this situation?

1. Find the total number of children: \(48 + 35 = 83\). 2. Divide by the number of children on each team: \(83 \div 9 = 9\) remainder \(2\). 3. The remainder is the number of children who help the referee, so \(2\) children help.

\(2\) children help the referee.

- First find the total number of rolls. - Divide the total by the number in each bag. - The remainder tells how many rolls are in the unfilled bag. - For the last part, find how many more are needed to make a group of eight.

1. Find the total number of rolls: \(56 + 65 = 121\). 2. Divide by the number of rolls in each bag: \(121 \div 8 = 15\) remainder \(1\). 3. The quotient shows that \(15\) bags can be filled completely. 4. The remainder shows that \(1\) roll is left over. 5. Find how many more rolls are needed for another full bag: \(8 - 1 = 7\).

a) He can fill \(15\) bags completely. b) \(1\) roll is left over. c) He would need to bake \(7\) more rolls.

- For each tray, find the greatest multiple of \(9\) that does not exceed the number of cookies. - Subtract that multiple from the number of cookies to find the remainder. - Compare the three remainders.

1. Find the remainder when the number of cookies on each tray is divided by \(9\). 2. Tray A: \(85 \div 9 = 9\) remainder \(4\). 3. Tray B: \(78 \div 9 = 8\) remainder \(6\). 4. Tray C: \(92 \div 9 = 10\) remainder \(2\). 5. Compare the remainders: \(2 < 4 < 6\). Tray C leaves the fewest cookies.

The baker should choose Tray C because it leaves only \(2\) cookies.

- Find the remainder for each possible group size. - Use nearby multiples of \(4\), \(5\), and \(6\). - Compare the three remainders.

1. Divide \(46\) by each possible group size and identify the remainder. 2. Groups of \(4\): \(46 \div 4 = 11\) remainder \(2\). 3. Groups of \(5\): \(46 \div 5 = 9\) remainder \(1\). 4. Groups of \(6\): \(46 \div 6 = 7\) remainder \(4\). 5. The smallest remainder is \(1\), so groups of \(5\) leave the fewest students without a complete group.

Groups of \(5\) leave the fewest students without a complete group.

- Find the greatest multiple of each divisor that does not exceed the dividend. - Subtract that multiple from the dividend to find the remainder. - Compare the four remainders.

1. For part a, \(7 \times 6 = 42\), and \(45 - 42 = 3\), so the result is \(7\) remainder \(3\). 2. For part b, \(8 \times 4 = 32\), and \(33 - 32 = 1\), so the result is \(8\) remainder \(1\). 3. For part c, \(7 \times 7 = 49\), and \(50 - 49 = 1\), so the result is \(7\) remainder \(1\). 4. For part d, \(8 \times 8 = 64\), and \(70 - 64 = 6\), so the result is \(8\) remainder \(6\). 5. The remainders are \(3\), \(1\), \(1\), and \(6\). The greatest remainder is \(6\).

a) \(7\) remainder \(3\) b) \(8\) remainder \(1\) c) \(7\) remainder \(1\) d) \(8\) remainder \(6\) Part d) has the greatest remainder, \(6\).

- Check whether the quotient and leftover amount reconstruct the dividend. - Ask whether any leftover amount could make another full group. - A check must satisfy both conditions, not just the total.

1. In a), \(7\times5+3=38\). The remainder \(3\) is smaller than \(5\), so the result is correct. 2. In b), \(6\times6+10=46\), but \(10\) leftovers can make another group of \(6\). So \(46\div6=7\) remainder \(4\). 3. In c), \(7\times8+3=59\). The remainder \(3\) is smaller than \(8\), so the result is correct.

a) Correct. b) Incorrect; \(46\div6=7\) remainder \(4\), because another full group of \(6\) could be made. c) Correct.

- Use multiplication as the inverse of division. - When the dividend is missing, multiply the quotient by the divisor and add the remainder. - When the divisor is missing, subtract the remainder before using the quotient. - Use \(\text{dividend} = \text{quotient} \times \text{divisor} + \text{remainder}\) to check each equation.

1. For part a, multiply the quotient by the divisor and add the remainder: \(5 \times 4 + 2 = 20 + 2 = 22\). 2. For part b, subtract the remainder first: \(29 - 1 = 28\). Then \(28 \div 4 = 7\), so the missing divisor is \(7\). 3. For part c, \(3 \times 8 + 5 = 24 + 5 = 29\). 4. For part d, \(7 \times 7 = 49\), and \(50 - 49 = 1\), so the quotient is \(7\) and the remainder is \(1\). 5. For part e, \(8 \times 5 = 40\), and \(43 - 40 = 3\), so the remainder is \(3\).

a) \(22\) b) \(7\) c) \(29\) d) \(7\) remainder \(1\) e) \(3\)

- Evaluate both quotients in each comparison. - Use related multiplication facts. - Compare the results.

1. a) \(150 \div 3 = 50\) and \(250 \div 5 = 50\), so the quotients are equal. 2. b) \(360 \div 4 = 90\) and \(450 \div 9 = 50\), so \(90 > 50\). 3. c) \(800 \div 2 = 400\) and \(1000 \div 2 = 500\), so \(400 < 500\). 4. d) \(210 \div 7 = 30\) and \(120 \div 4 = 30\), so the quotients are equal. 5. e) \(540 \div 6 = 90\) and \(720 \div 8 = 90\), so the quotients are equal.

a) \(=\) b) \(>\) c) \(<\) d) \(=\) e) \(=\)

- Divide \(420\) by each bag size. - Compare the two quotients. - Think about what happens to the number of bags when each bag holds more apples.

1. For part a, divide by \(6\): \(420 \div 6 = 70\). 2. For part b, divide by \(7\): \(420 \div 7 = 60\). 3. Compare the results: \(60 < 70\), so part b uses \(10\) fewer bags. 4. When each bag holds more apples, fewer bags are needed for the same total number of apples.

a) \(70\) bags are needed. b) \(60\) bags are needed. c) Part b needs fewer bags because each bag holds more apples.

- Find how many full groups of the divisor fit in each dividend. - Subtract the corresponding multiple from the dividend to find the remainder. - Compare each remainder with \(4\).

1. For part a, \(5 \times 5 = 25\), and \(29 - 25 = 4\), so the result is \(5\) remainder \(4\). 2. For part b, \(4 \times 8 = 32\), and \(38 - 32 = 6\), so the result is \(4\) remainder \(6\). 3. For part c, \(6 \times 7 = 42\), and \(46 - 42 = 4\), so the result is \(6\) remainder \(4\). 4. For part d, \(3 \times 9 = 27\), and \(31 - 27 = 4\), so the result is \(3\) remainder \(4\). 5. For part e, \(8 \times 6 = 48\), and \(50 - 48 = 2\), so the result is \(8\) remainder \(2\). 6. Parts a, c, and d have remainder \(4\).

a) \(5\) remainder \(4\) b) \(4\) remainder \(6\) c) \(6\) remainder \(4\) d) \(3\) remainder \(4\) e) \(8\) remainder \(2\) Parts a), c), and d) have remainder \(4\).

- List the possible remainder values for each divisor before choosing the largest one. - Multiply the quotient by the divisor after choosing the remainder. - Add the remainder to that product to reconstruct the dividend.

1. A remainder must be less than the divisor, so choose the greatest allowed remainder in each case. 2. a) Remainder \(3\): \(8\times4+3=35\). 3. b) Remainder \(6\): \(5\times7+6=41\). 4. c) Remainder \(8\): \(6\times9+8=62\). 5. d) Remainder \(5\): \(7\times6+5=47\).

a) \(35\div4=8\) remainder \(3\) b) \(41\div7=5\) remainder \(6\) c) \(62\div9=6\) remainder \(8\) d) \(47\div6=7\) remainder \(5\)

- Notice how the dividend changes from one problem to the next. - Track how each remainder changes. - Look at what happens when the dividend is exactly divisible by \(4\).

1. The results are: a) \(4\) remainder \(2\); b) \(4\) remainder \(3\); c) \(5\) remainder \(0\); d) \(5\) remainder \(1\); e) \(5\) remainder \(2\). 2. As the dividend increases by \(1\), the remainder also increases by \(1\). After reaching the greatest possible remainder, \(3\), the next dividend is a multiple of \(4\). The quotient increases by \(1\), and the remainder starts again at \(0\).

a) \(4\) remainder \(2\) b) \(4\) remainder \(3\) c) \(5\) remainder \(0\) d) \(5\) remainder \(1\) e) \(5\) remainder \(2\) The remainder increases by \(1\) as the dividend increases by \(1\). After remainder \(3\), it starts again at \(0\), and the quotient increases by \(1\).

- Work out all four quotients before ordering them. - Do not compare the dividends alone: the divisors are different. - Compare the hundreds digits of the quotients first, then the tens and ones if needed.

1. \(812\div4=203\). 2. \(945\div7=135\). 3. \(585\div5=117\). 4. \(756\div6=126\). 5. From least to greatest, \(117<126<135<203\), so the order is C, D, B, A.

A) \(203\) B) \(135\) C) \(117\) D) \(126\) Order: C, D, B, A.

- Use multiplication to reverse division. - To reconstruct a dividend, multiply divisor and quotient, then add the remainder. - For a missing one-digit divisor, look for the one-digit factor that multiplies the quotient to make the dividend.

1. a) Reconstruct the dividend: \(123\times4+2=494\). 2. b) \(105\times7=735\), so the remainder is \(739-735=4\). 3. c) Find the one-digit factor that makes \(141\times\square=846\). Since \(141\times6=846\), the missing divisor is \(6\).

a) \(494\) b) \(4\) c) \(6\)

- Calculate all four quotients first. - Divide from left to right using the standard algorithm. - Compare the quotient values by place value.

1. \(534 \div 3=178\). 2. \(968 \div 8=121\). 3. \(745 \div 5=149\). 4. \(864 \div 6=144\). 5. Since \(178>149>144>121\), part a) has the greatest quotient.

a) \(178\) b) \(121\) c) \(149\) d) \(144\) The greatest quotient is \(178\) in part a).

- Identify the two parts that make the original dividend. - Use multiplication to work backward when a partial dividend is missing. - Add the partial quotients.

1. For part a, \(80 \div 4 = 20\) and \(4 \div 4 = 1\). Then \(20 + 1 = 21\). 2. For part b, the missing partial dividend is \(90\), because \(90 \div 3 = 30\). Also, \(6 \div 3 = 2\). Then \(30 + 2 = 32\). 3. For part c, \(50 \div 5 = 10\) and \(25 \div 5 = 5\). Then \(10 + 5 = 15\).

a) \(80 \div 4 = 20\), \(4 \div 4 = 1\), so \(84 \div 4 = 21\) b) \(90 \div 3 = 30\), \(6 \div 3 = 2\), so \(96 \div 3 = 32\) c) \(50 \div 5 = 10\), \(25 \div 5 = 5\), so \(75 \div 5 = 15\)

- Write two partial-division steps for each problem. - Make the first partial dividend ten times the divisor. - Divide the remaining part by the same divisor. - Add the partial quotients and compare the results.

1. For part a, \(40 \div 4 = 10\) and \(12 \div 4 = 3\). Then \(10 + 3 = 13\). 2. For part b, \(60 \div 6 = 10\) and \(18 \div 6 = 3\). Then \(10 + 3 = 13\). 3. For part c, \(70 \div 7 = 10\) and \(21 \div 7 = 3\). Then \(10 + 3 = 13\). 4. For part d, \(80 \div 8 = 10\) and \(24 \div 8 = 3\). Then \(10 + 3 = 13\). 5. All four quotients are equal.

a) \(13\) b) \(13\) c) \(13\) d) \(13\) All four quotients are equal.

- Divide the second number by the first number. - Decompose the dividend into parts that are easy to divide. - For \(114 \div 3\), begin with a large multiple of \(3\) that you know.

1. For a), compute \(114 \div 3\). Decompose \(114\) as \(90 + 24\): \(90 \div 3 = 30\) and \(24 \div 3 = 8\). Then \(30 + 8 = 38\). 2. For b), compute \(153 \div 9\). Decompose \(153\) as \(90 + 63\): \(90 \div 9 = 10\) and \(63 \div 9 = 7\). Then \(10 + 7 = 17\).

a) \(114\) is \(38\) times as great as \(3\). b) \(153\) is \(17\) times as great as \(9\).

- First calculate the exact number of trays needed. - Break \(135\) into numbers that are easy to divide by \(9\). - Compare your result with \(20\).

1. Find the number of trays needed: \(135 \div 9\). 2. Break \(135\) into compatible numbers: \(135 = 90 + 45\). 3. Divide each part: \(90 \div 9 = 10\) and \(45 \div 9 = 5\). Then add: \(10 + 5 = 15\). 4. Only \(15\) trays are needed, so the statement that at least \(20\) are needed is false.

No. Only \(15\) trays are needed because \(135 \div 9 = 15\).

- Break \(192\) into two multiples of \(6\). - First find the number of pots on one table. - Then divide that amount equally among \(4\) children.

1. Divide the pots among the tables: \(192 \div 6 = (180 \div 6) + (12 \div 6) = 30 + 2 = 32\). 2. Divide one table’s \(32\) pots among \(4\) children: \(32 \div 4 = 8\).

Each table has \(32\) pots. Each of the \(4\) children gets \(8\) pots.

- Find the remainder when each number is divided by \(7\). - Use the greatest multiple of \(7\) that does not exceed each number. - Compare the three remainders.

1. Find the remainder for the tulips: \(80 \div 7 = 11\) remainder \(3\). 2. Find the remainder for the daffodils: \(95 \div 7 = 13\) remainder \(4\). 3. Find the remainder for the crocuses: \(110 \div 7 = 15\) remainder \(5\). 4. Compare the remainders: \(5 > 4 > 3\). The crocuses leave the greatest number of plants.

The crocuses leave the most plants, with \(5\) left over.

- How many apples are there altogether? - Determine how many complete groups of eight can be made. - What does the remainder represent in this situation?

1. Find the total number of apples: \(46 + 35 = 81\). 2. Divide by the number of apples in each bag: \(81 \div 8 = 10\) remainder \(1\). 3. The remainder shows that \(1\) apple is left over.

\(1\) apple is left over.

- First find the total number of children. - Divide by each proposed group size and identify the remainder. - Compare the two remainders.

1. The total is \(28+25=53\) children. 2. \(53\div6=8\) remainder \(5\). 3. \(53\div5=10\) remainder \(3\). 4. Since \(3<5\), groups of \(5\) leave fewer children left over.

a) \(5\) children b) \(3\) children c) Groups of \(5\)

- Find the greatest multiple of each divisor that does not exceed the dividend. - Subtract to find each remainder. - Compare the remainders, then subtract the smaller remainder from the greater one.

1. For part a, \(9 \times 8 = 72\), and \(74 - 72 = 2\), so the result is \(9\) remainder \(2\). 2. For part b, \(9 \times 6 = 54\), and \(59 - 54 = 5\), so the result is \(9\) remainder \(5\). 3. Since \(5 > 2\), part b has the greater remainder. 4. The difference between the remainders is \(5 - 2 = 3\).

a) \(9\) remainder \(2\) b) \(9\) remainder \(5\) Part b) has the greater remainder. The difference between the remainders is \(3\).

- Write multiples of the divisor around the given range. - What must you add to each multiple to leave the required remainder? - Check the lower and upper limits of the range.

1. Start with multiples of \(7\) near the interval: \(28,35,42,49\). 2. A number that leaves remainder \(4\) is \(4\) more than a multiple of \(7\). The candidates are \(32,39,46,53\). 3. The numbers from \(30\) through \(50\) are \(32,39,46\). 4. Divide to check: \(32\div7=4\) remainder \(4\), \(39\div7=5\) remainder \(4\), and \(46\div7=6\) remainder \(4\).

\(32\div7=4\) remainder \(4\) \(39\div7=5\) remainder \(4\) \(46\div7=6\) remainder \(4\)

- Work backward by multiplying the quotient by the divisor. - Add the remainder to find the dividend. - List the possible remainders when the divisor is \(7\).

1. For part a, multiply the quotient by the divisor and add the remainder: \(13 \times 7 + 5 = 91 + 5 = 96\). The mystery number is \(96\). 2. For part b, a remainder must be less than the divisor. The greatest whole number less than \(7\) is \(6\), so the greatest possible remainder is \(6\).

a) The mystery number is \(96\). b) The greatest possible remainder is \(6\), because a remainder must be less than the divisor \(7\).

- Look for a large multiple of \(9\) close to \(855\). - Think about how a useful decomposition can turn one difficult division into easier familiar facts. - Explain why each chosen part works cleanly with the divisor.

1. Divide each part by \(9\): \(810 \div 9 = 90\) and \(45 \div 9 = 5\). 2. Add the partial quotients: \(90 + 5 = 95\). Each bed gets \(95\) seedlings. 3. The decomposition is efficient because both \(810\) and \(45\) are familiar multiples of \(9\) and divide evenly by \(9\).

a) Each bed gets \(95\) seedlings. b) The decomposition is efficient because both \(810\) and \(45\) are familiar multiples of \(9\).

- What must always be true about a remainder compared with its divisor? - Use the fact that \(5\) is allowed but \(6\) is not. - For part b, start with a multiple of the divisor and adjust it to leave remainder \(3\).

1. A valid remainder must be less than the divisor. Since \(5\) is possible but \(6\) is not, the divisor is \(6\). 2. For part b, choose a multiple of \(6\) and add \(3\). For example, \(2\times6+3=15\), so \(15\div6=2\) remainder \(3\).

a) The divisor is \(6\). b) One possible example is \(15\div6=2\) remainder \(3\).

- Calculate both quotients in each comparison. - Compare the exact quotient values from left to right by place value. - Recall the meanings of \(<\), \(>\), and \(=\).

1. For a), \(3216 \div 4=804\) and \(4824 \div 6=804\), so the quotients are equal. 2. For b), \(2555 \div 5=511\) and \(1836 \div 3=612\), so the left quotient is less.

a) \(=\) b) \(<\)

- Complete each division through the final remainder. - Include any required zero in a quotient. - Compare the three final remainders.

1. \(458 \div 6=76\text{ R }2\). 2. \(723 \div 7=103\text{ R }2\). 3. \(914 \div 4=228\text{ R }2\). 4. All three divisions have the same remainder, \(2\).

a) \(76\text{ R }2\) b) \(103\text{ R }2\) c) \(228\text{ R }2\) All three remainders are \(2\).

- Divide the first bill by the price per tree. - Subtract the delivered number from the total needed. - Multiply the missing number of trees by the unit price.

1. The first delivery contains \(\$1456 \div \$4 = 364\) trees. 2. The park still needs \(400 - 364 = 36\) trees. 3. The additional trees cost \(36 \times \$4 = \$144\).

a) The first delivery contains \(364\) trees. b) The city must pay \(\$144\) more.

- Calculate all four quotients. - Compare the final values. - Recheck the expression whose quotient appears different.

1. \(1344 \div 3=448\). 2. \(1792 \div 4=448\). 3. \(2688 \div 6=448\). 4. \(3592 \div 8=449\). 5. Part d) has the different quotient.

d) \(3592 \div 8=449\) has a different quotient. The other three quotients are \(448\).

- Calculate each quotient separately. - Compare the quotient values by place value. - The values differ only in the ones place.

1. \(6552 \div 7=936\). 2. \(8433 \div 9=937\). 3. Since \(936<937\), the correct symbol is \(<\).

\(6552 \div 7<8433 \div 9\) because \(936<937\).

- Look for the same three numbers in each column. - Solving the division equation in the second column gives the missing factor above it. - In a related division equation, the product becomes the dividend.

1. In the first column, \(417\times 3=1251\). The related division equation is \(1251\div 3=417\). 2. In the second column, divide to find the missing factor: \(1345\div 5=269\). 3. Use that value in the related multiplication equation: \(269\times 5=1345\).

First column: \(417\times 3=1251\) and \(1251\div 3=417\). Second column: \(269\times 5=1345\) and \(1345\div 5=269\).

- Follow the place values from left to right. - Ask how many times the divisor fits into the amount at each stage. - Preserve the quotient place with a zero when the answer is zero times.

1. For a), \(8 \div 4=2\). Bring down the \(1\). Since \(4\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(6\) to make \(16\), and \(16 \div 4=4\). Therefore, \(816 \div 4=204\). 2. For b), \(9 \div 3=3\). Bring down the \(2\). Since \(3\) goes into \(2\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(1\) to make \(21\), and \(21 \div 3=7\). Therefore, \(921 \div 3=307\). 3. In each division, the digit brought down for the tens place is less than the divisor. The divisor fits zero times in that place, so a zero must be written in the tens place of the quotient before continuing.

a) \(204\) b) \(307\) c) The zero is required because \(1<4\) and \(2<3\), so the divisor fits zero times in the tens place.

- Check Niran's quotient by multiplying it by \(6\) and comparing with the dividend. - In the standard algorithm, keep the quotient digits aligned with the dividend's place values. - When a partial dividend is too small to make a nonzero quotient digit, consider what must be recorded to preserve place value.

1. Part a: Since \(4<6\), use the first two digits. \(48\div 6=8\), so write \(8\) in the hundreds place. Bring down \(3\). Because \(3<6\), write \(0\) in the tens place. Bring down \(6\), and \(36\div 6=6\). Therefore, \(4836\div 6=806\). 2. Niran probably omitted the zero in the tens place and wrote \(86\) instead of \(806\). 3. Part c: \(7\div 7=1\), so write \(1\) in the thousands place. Bring down \(0\) and write \(0\) in the hundreds place. Bring down \(5\) and write \(0\) in the tens place because \(5<7\). Bring down \(6\), and \(56\div 7=8\). Therefore, \(7056\div 7=1008\).

a) \(806\) b) Niran omitted the zero in the tens place. c) \(1008\)

- Estimate how many digits the quotient should have. - In the standard algorithm, write a zero in the quotient when a partial dividend is smaller than the divisor. - Multiply the quotient by \(5\) to check your result.

1. Round \(2015\) to \(2000\). Then \(2000\div 5=400\). Since \(43\) is much smaller than \(400\), Max’s answer cannot be correct. 2. Begin with \(20\) because \(2<5\). Then \(20\div 5=4\). Bring down the \(1\). Since \(5\) goes into \(1\) zero times, write \(0\) in the tens place of the quotient. Bring down the \(5\) to make \(15\), and \(15\div 5=3\). Therefore, \(2015\div 5=403\). 3. Check: \(403\times 5=2015\).

1. One estimate is \(2000\div 5=400\), so \(43\) is much too small. 2. The correct quotient is \(403\). 3. Check: \(403\times 5=2015\).

- Divide the product by the known factor to find the missing factor. - Use multiplication to check a division result. - Keep track of each remainder as you divide from left to right.

1. Divide the product by the known factor: \(8136\div 8\). 2. Since \(8\div 8=1\), write \(1\). Bring down the \(1\). Since \(8\) goes into \(1\) zero times, write \(0\) in the hundreds place of the quotient. Bring down the \(3\) to make \(13\), and \(13\div 8=1\) with remainder \(5\). Bring down the \(6\) to make \(56\), and \(56\div 8=7\). Therefore, \(8136\div 8=1017\). 3. Check by multiplication: \(1017\times 8=8136\). The product matches the given value.

a) The unknown number is \(1017\). b) Check: \(1017\times 8=8136\).

- Start with a large familiar multiple of \(7\). - Break the remaining amount into another multiple of \(7\). - Add the partial quotients.

1. Break \(861\) into \(700 + 140 + 21\), all multiples of \(7\). 2. Divide each part: \(700 \div 7 = 100\), \(140 \div 7 = 20\), and \(21 \div 7 = 3\). 3. Add the partial quotients: \(100 + 20 + 3 = 123\).

Each box contains \(123\) posters.

- Divide \(180\) by each number of books per section. - Compare the two quotients. - Explain how increasing the group size changes the number of groups.

1. For part a, divide: \(180 \div 6 = 30\). 2. For part b, divide: \(180 \div 9 = 20\). 3. Since each section holds more books in part b, the same total number of books is divided into fewer groups.

a) \(30\) shelf sections are needed. b) \(20\) shelf sections are needed. c) The second arrangement uses fewer sections because each section holds more books.

- Use related facts with \(45\). - For the same dividend, compare how the divisor affects the quotient. - Check each quotient with multiplication.

1. a) The related fact \(45 \div 5 = 9\) gives \(450 \div 5 = 90\). 2. b) With the same dividend, dividing by the greater divisor gives the smaller quotient, so \(450 \div 9 < 450 \div 5\). 3. c) The related fact \(45 \div 9 = 5\) gives \(450 \div 9 = 50\).

a) \(90\); related fact: \(45 \div 5 = 9\) b) Less than, because the divisor is greater. c) \(50\)

- Use multiplication to find a missing dividend. - For a missing one-digit divisor, look for the factor that multiplies the quotient to make the dividend. - Break a dividend into convenient parts when finding a quotient.

1. In a), \(120\times4=480\), so the missing dividend is \(480\). 2. In b), find the one-digit factor that makes \(70\times\square=560\). Since \(70\times8=560\), the missing divisor is \(8\). 3. In c), \(840\div2=420\). 4. In d), \(210\times3=630\), so the missing dividend is \(630\). 5. In e), \(500\div5=100\) and \(150\div5=30\), so \(650\div5=130\).

a) \(480\) b) \(8\) c) \(420\) d) \(630\) e) \(130\)

- Break \(840\) into numbers that are easy to divide by \(6\). - Add the partial quotients for part a. - For part b, compare the two carton sizes without finding the exact second quotient.

1. For part a, divide \(840\) by \(6\). Break apart \(840 = 600 + 240\). 2. Divide each part: \(600 \div 6 = 100\) and \(240 \div 6 = 40\). Then add: \(100 + 40 = 140\). 3. For part b, cartons holding \(3\) eggs are half the size of cartons holding \(6\). Smaller groups mean more groups are needed for the same total.

a) The farmer can fill \(140\) cartons. b) More cartons would be needed because each carton holds fewer eggs.

- Use multiplication and addition to work backward when the dividend is missing. - When the divisor is missing, subtract the remainder first. - A remainder is the difference between the dividend and the product of the quotient and divisor.

1. For part a, multiply the quotient by the divisor and add the remainder: \(7 \times 6 + 3 = 42 + 3 = 45\). 2. For part b, subtract the remainder: \(38 - 3 = 35\). Then divide by the quotient: \(35 \div 5 = 7\), so the missing divisor is \(7\). 3. For part c, \(6 \times 8 = 48\), and \(50 - 48 = 2\), so the missing remainder is \(2\).

a) \(45\) b) \(7\) c) \(2\)

- How many children are going on the trip altogether? - How many complete groups of six can be formed? - Do the children in the remainder also need a guide? - Include every group when finding the number of guides.

1. Find the total number of children: \(68 + 75 + 82 + 71 = 296\). 2. Divide by the maximum number in each group: \(296 \div 6 = 49\) remainder \(2\). 3. The remaining two children need another group and guide, so \(49 + 1 = 50\) guides are needed.

The schools need \(50\) museum guides.

- Use multiplication to work backward from the quotient and divisor. - After finding the amount in complete groups, account for the remainder. - Check that your reconstructed dividend gives quotient \(8\) and remainder \(5\) when divided by \(7\).

1. Multiply the quotient by the divisor: \(8\times7=56\). 2. Add the remainder: \(56+5=61\). 3. The missing number is \(61\).

\(61\)

- Break \(672\) into familiar multiples of \(6\). - Divide each part by \(6\). - Add the partial quotients.

1. Break \(672\) into \(600 + 60 + 12\), all divisible by \(6\). 2. Divide each part: \(600 \div 6 = 100\), \(60 \div 6 = 10\), and \(12 \div 6 = 2\). 3. Add the partial quotients: \(100 + 10 + 2 = 112\).

A total of \(112\) cases are delivered.

- First find the number of beads on one necklace for each group. - Break \(192\) and \(234\) into parts that are easy to divide by \(4\) and \(6\). - Compare the two results.

1. Find the number of beads on each Group A necklace: \(192 \div 4 = 48\). 2. Find the number of beads on each Group B necklace: \(234 \div 6 = 39\). 3. Since \(48 > 39\), Group A puts more beads on each necklace.

Group A puts more beads on each necklace: \(48\) beads compared with Group B's \(39\) beads.

- Find both quotients in each part. - Record intermediate results if helpful. - Compare the final values.

1. For part a, \(480 \div 4 = 120\) and \(360 \div 3 = 120\), so the quotients are equal. 2. For part b, \(505 \div 5 = 101\) and \(808 \div 8 = 101\), so the quotients are equal. 3. For part c, \(609 \div 3 = 203\) and \(406 \div 2 = 203\), so the quotients are equal. 4. For part d, \(903 \div 3 = 301\) and \(804 \div 2 = 402\), so \(301 < 402\).

a) \(=\) b) \(=\) c) \(=\) d) \(<\)

- Calculate each quotient first. - Compare the place of the zero in each pair. - Identify which digits remain the same.

1. a) \(360 \div 3 = 120\) and \(306 \div 3 = 102\). 2. b) \(840 \div 4 = 210\) and \(804 \div 4 = 201\). 3. c) \(550 \div 5 = 110\) and \(505 \div 5 = 101\). 4. In each pair, the same nonzero digits appear, while the zero shifts from the ones place to the tens place in the quotient.

a) \(120\) and \(102\) b) \(210\) and \(201\) c) \(110\) and \(101\) In each pair, the zero and the final nonzero digit switch places in the quotient.

- Compare each new divisor or dividend with the given equation. - Think about sharing the same amount among twice as many groups. - Think about sharing half as much among the same number of groups.

1. In a), the dividend stays fixed and the divisor doubles from \(3\) to \(6\), so the quotient is halved: \(300 \div 2 = 150\). 2. In b), the divisor stays fixed and the dividend is halved from \(900\) to \(450\), so the quotient is also halved: \(300 \div 2 = 150\).

a) \(150\) b) \(150\) Doubling the divisor halves the quotient in a); halving the dividend halves the quotient in b).

- Find each quotient separately. - Use place-value parts to divide by \(3\). - Compare the hundreds, tens, and ones in the quotients.

1. The quotients are \(630 \div 3 = 210\), \(360 \div 3 = 120\), \(603 \div 3 = 201\), and \(306 \div 3 = 102\). 2. In increasing order, \(102 < 120 < 201 < 210\), so the order is d, b, c, a.

a) \(630 \div 3 = 210\) b) \(360 \div 3 = 120\) c) \(603 \div 3 = 201\) d) \(306 \div 3 = 102\) From smallest quotient to greatest quotient: d), b), c), a).

- Divide \(168\) by each team size. - Use a decomposition that works well with each divisor. - Compare the two quotients.

1. For teams of \(6\), break \(168\) into compatible numbers: \(168 = 120 + 48\). Then \(120 \div 6 = 20\) and \(48 \div 6 = 8\), so \(20 + 8 = 28\) teams. 2. For teams of \(4\), use \(168 = 160 + 8\). Then \(160 \div 4 = 40\) and \(8 \div 4 = 2\), so \(40 + 2 = 42\) teams. 3. Since \(42 > 28\), the smaller team size creates more teams.

Teams of \(6\) produce \(28\) teams. Teams of \(4\) produce \(42\) teams, so the smaller team size creates more teams.

- Find a large multiple of \(8\) close to \(744\). - Choose a different large multiple of \(8\) for the second method. - Make sure every part is divisible by \(8\).

1. One decomposition is \(744 = 720 + 24\). Then \(720 \div 8 = 90\) and \(24 \div 8 = 3\), so the quotient is \(93\). 2. Another decomposition is \(744 = 640 + 104\). Then \(640 \div 8 = 80\) and \(104 \div 8 = 13\). The second partial quotient can be found from \(104 = 80 + 24\): \(80 \div 8 = 10\) and \(24 \div 8 = 3\). Thus \(80 + 13 = 93\).

The quotient is \(93\). One method uses \(720 \div 8 = 90\) and \(24 \div 8 = 3\). Another method uses \(640 \div 8 = 80\) and \(104 \div 8 = 13\).

- Calculate both quotients in each pair. - If the dividends are equal but divisors differ, think about the size of each share. - Check close quotients using multiplication.

1. In a), \(612\div6=102\) and \(618\div6=103\), so \(618\div6\) is greater. 2. In b), \(832\div8=104\) and \(832\div4=208\), so \(832\div4\) is greater. 3. In c), \(918\div9=102\) and \(412\div4=103\), so \(412\div4\) is greater.

a) \(612\div6=102\), \(618\div6=103\); \(618\div6\) is greater. b) \(832\div8=104\), \(832\div4=208\); \(832\div4\) is greater. c) \(918\div9=102\), \(412\div4=103\); \(412\div4\) is greater.

- Break \(468\) into parts that are easy to divide by \(4\). - Divide each part by \(4\). - Add the partial quotients.

1. Break \(468\) into \(400 + 60 + 8\), with each part divisible by \(4\). 2. Divide each part: \(400 \div 4 = 100\), \(60 \div 4 = 15\), and \(8 \div 4 = 2\). 3. Add the partial quotients: \(100 + 15 + 2 = 117\).

Each child receives \(\$117\).

- Look for what stays the same in each pair. - With the same dividend, think about how the divisor affects the quotient. - With the same divisor, think about how the dividend affects the quotient. - In part c), compare how both numbers change.

1. In part a), the dividend is the same. Dividing by the smaller divisor gives the greater quotient: \(400 \div 5 = 80\) and \(400 \div 8 = 50\), so \(80 > 50\). 2. In part b), the divisor is the same. The greater dividend gives the greater quotient: \(240 \div 4 = 60\) and \(480 \div 4 = 120\), so \(60 < 120\). 3. In part c), both the dividend and divisor are doubled: \(180 \times 2 = 360\) and \(2 \times 2 = 4\). The quotient stays the same, and \(180 \div 2 = 360 \div 4 = 90\).

a) \(>\) b) \(<\) c) \(=\); both the dividend and divisor are doubled, so the quotient stays the same.

- Look for familiar multiples of \(6\) or \(60\). - Make sure the parts add to \(876\). - Divide each part and add the partial quotients.

1. Using Lukas’s method, \(600 \div 6 = 100\), \(240 \div 6 = 40\), and \(36 \div 6 = 6\). Then \(100 + 40 + 6 = 146\). 2. One alternative is \(876 = 840 + 36\). Then \(840 \div 6 = 140\) and \(36 \div 6 = 6\), so \(140 + 6 = 146\).

a) \(146\) b) One example is \(840 \div 6 = 140\) and \(36 \div 6 = 6\), so the quotient is \(146\).

- Use the multiplication check for division with a remainder. - In part b, subtract the remainder before finding the divisor. - Check that the remainder is less than the divisor you find.

1. For part a, use \(\text{dividend} = \text{quotient} \times \text{divisor} + \text{remainder}\): \(7 \times 6 + 4 = 42 + 4 = 46\). 2. For part b, subtract the remainder first: \(29 - 2 = 27\). Then divide by the quotient: \(27 \div 9 = 3\). The divisor is \(3\).

a) The dividend is \(46\). b) The divisor is \(3\).

- Break \(156\) into two multiples of \(6\). - Divide each part by \(6\). - Add the partial quotients.

1. The division problem is \(156 \div 6\). 2. Break \(156\) into \(120 + 36\). 3. Divide each part: \(120 \div 6 = 20\) and \(36 \div 6 = 6\). 4. Add the partial quotients: \(20 + 6 = 26\).

The gardener can plant \(26\) rows.

- First calculate the exact number of cases needed. - Break \(336\) into multiples of \(8\). - Compare the quotient with \(40\).

1. Find the number of cases needed: \(336 \div 8\). 2. Break \(336\) into compatible numbers: \(336 = 320 + 16\). 3. Divide each part: \(320 \div 8 = 40\) and \(16 \div 8 = 2\). 4. Add: \(40 + 2 = 42\). Since \(42 \ne 40\), the teacher is not correct.

No. The trip needs \(42\) cases, so \(40\) cases are not enough.

- Test each student's result by rebuilding the original total. - For each leftover amount, ask whether another full group can be made. - Different errors can produce a wrong quotient and remainder.

1. For Lucas, \(8\times9+4=76\), and \(4<9\); his result is valid. 2. For Mia, \(7\times9+13=76\), but \(13\) leftovers can form another group of \(9\). So her quotient is too small. 3. For Sophie, \(8\times9+3=75\), not \(76\); her result does not reconstruct the dividend.

Lucas is correct. Mia is incorrect because \(13\) leftovers could make another group of \(9\). Sophie is incorrect because \(8\times9+3=75\), not \(76\).

- Compare the thousands digit with the divisor \(6\). - Decide whether division begins with the thousands digit or with the first two digits. - Choose one four-digit dividend on each side of that threshold.

1. For a), choose a four-digit number with a thousands digit less than \(6\), such as \(1200\). Then \(1200 \div 6=200\), which has \(3\) digits. 2. For b), choose a four-digit number with a thousands digit at least \(6\), such as \(6000\). Then \(6000 \div 6=1000\), which has \(4\) digits. 3. The thousands digit must be at least \(6\).

a) For example, \(1200 \div 6=200\). b) For example, \(6000 \div 6=1000\). c) ... at least \(6\).

- Complete part a) with the standard division algorithm. - Compare \(2688\) with \(1344\). - Decide how the same scale factor affects the quotient when the divisor is unchanged.

1. Part a: Since \(1<6\), use \(13\). Then \(13\div 6=2\) remainder \(1\). Bring down \(4\): \(14\div 6=2\) remainder \(2\). Bring down the final \(4\): \(24\div 6=4\). Therefore, \(1344\div 6=224\). 2. The dividend \(2688\) is twice \(1344\). 3. With the divisor unchanged, doubling the dividend doubles the quotient, so \(224\times 2=448\).

a) \(224\) b) \(448\). The dividend is doubled while the divisor stays the same, so the quotient is doubled.

- First find how many pounds of paper are in one container for each class. - Break each dividend into hundreds and tens that are easy to divide by the given divisor. - Compare the two amounts per container.

1. Find the amount in each container for Ms. Green's class: \(840\,\text{lb} \div 4 = (800\,\text{lb} \div 4) + (40\,\text{lb} \div 4) = 200\,\text{lb} + 10\,\text{lb} = 210\,\text{lb}\). 2. Find the amount in each container for Mr. Lee's class: \(750\,\text{lb} \div 3 = (600\,\text{lb} \div 3) + (150\,\text{lb} \div 3) = 200\,\text{lb} + 50\,\text{lb} = 250\,\text{lb}\). 3. Compare the amounts: \(250\,\text{lb} > 210\,\text{lb}\).

Mr. Lee's class has more paper in each container, with \(250\,\text{lb}\) compared with \(210\,\text{lb}\) for Ms. Green's class.

- First calculate the side with no missing number. - Both sides of each equation must have the same value. - Use inverse multiplication to find a missing one-digit divisor.

1. For a), \(324\div4=81\). The missing dividend must satisfy \(\square\div9=81\), so \(81\times9=729\). 2. For b), \(735\div7=105\). The missing divisor must satisfy \(105\times\square=525\). Since \(105\times5=525\), the missing divisor is \(5\).

a) \(729\) b) \(5\)

- Find the greatest multiple of the divisor that does not exceed the dividend. - Subtract that multiple from the dividend to find the remainder. - Check that the remainder is less than the divisor.

1. \(6 \times 24 = 144\), and \(148 - 144 = 4\). Therefore, \(148 \div 6 = 24\text{ R }4\). 2. \(255 \div 5 = 51\) with no remainder. 3. \(8 \times 39 = 312\), and \(314 - 312 = 2\). Therefore, \(314 \div 8 = 39\text{ R }2\). 4. \(7 \times 142 = 994\), and \(1000 - 994 = 6\). Therefore, \(1000 \div 7 = 142\text{ R }6\).

a) \(24\text{ R }4\) b) \(51\) c) \(39\text{ R }2\) d) \(142\text{ R }6\)

- Find the quotient and remainder for each division. - A remainder of \(3\) means the dividend is \(3\) greater than a multiple of the divisor. - Verify that each remainder is less than its divisor.

1. \(59 \div 8 = 7\text{ R }3\) because \(8 \times 7 = 56\). 2. \(115 \div 4 = 28\text{ R }3\) because \(4 \times 28 = 112\). 3. \(83 \div 5 = 16\text{ R }3\) because \(5 \times 16 = 80\). 4. \(200 \div 6 = 33\text{ R }2\) because \(6 \times 33 = 198\). 5. Parts a), b), and c) have remainder \(3\).

a) \(7\text{ R }3\) b) \(28\text{ R }3\) c) \(16\text{ R }3\) d) \(33\text{ R }2\) Parts a), b), and c) have remainder \(3\).

- For each estimate, choose a nearby dividend that is easy to divide by the divisor. - In the standard algorithm, keep every place represented in the quotient, even when a partial dividend is too small to contain the divisor. - Check each exact quotient by multiplication.

1. Estimate \(4800\div8=600\). Using the standard algorithm, \(4656\div8=582\). 2. Estimate \(6000\div6=1000\). Using the standard algorithm, \(6216\div6=1036\). The \(0\) in the hundreds place of the quotient is needed because after dividing \(6\) thousands, the next partial dividend is \(2\), which is less than \(6\).

1) Estimate: \(600\); exact quotient: \(582\) 2) Estimate: \(1000\); exact quotient: \(1036\)

- Count the dots in one complete row to find the bag size. - Count only complete rows as filled bags. - Compare the dots left in the short row with the full-row size.

1. The array shows \(29\) marbles and \(4\) dots in each full row. 2. Seven rows use \(7\times4=28\) marbles, leaving \(1\). 3. One marble is not enough for another bag of \(4\). Thus \(29\div4=7\) remainder \(1\).

\(7\) full bags with \(1\) marble left; no, there are not enough for another full bag.

- Count the complete rows and the dots in each. - Count the dots in the final, incomplete row separately. - Check that the quotient and remainder account for every dot.

1. The array shows \(45\) dots with \(7\) in each complete row. 2. Six full rows use \(6\times7=42\) dots, leaving \(45-42=3\). 3. Noah counted one too few leftover dots. The correct equation is \(45\div7=6\) remainder \(3\).

Noah missed one leftover dot. \(45\div7=6\) remainder \(3\)

- Count the dots in a full row before judging the claim. - Could all the claimed leftover dots remain if you continued making rows? - Find the actual short row in the diagram.

1. The array contains \(50\) dots and has \(8\) dots in each complete row. 2. Six complete rows contain \(6\times8=48\) dots, leaving \(2\). Thus \(50\div8=6\) remainder \(2\). 3. Nine leftover dots are impossible after making all full rows: eight of them would form another full row.

No. There are \(6\) full rows and \(2\) dots left over. \(50\div8=6\) remainder \(2\).

- Count the uncolored dots, or subtract the colored dots from the total. - Divide the number of good eggs by the number that fit in one carton. - Interpret the quotient and remainder in the question.

1. Count \(50\) dots, including \(7\) colored dots. 2. Good eggs: \(50-7=43\). 3. Divide into cartons: \(43\div6=7\) remainder \(1\). 4. There are \(7\) full cartons and \(1\) good egg left over.

\(7\) full cartons; \(1\) good egg left over.

- Count the dots in each full row to find the capacity of a carton. - Count the full rows and then the incomplete one. - How many dots would complete the last row?

1. The diagram shows \(60\) dots with \(8\) in each complete row. 2. Seven full rows account for \(7\times8=56\) lunch boxes, leaving \(4\). 3. Another carton holds \(8\), so it needs \(8-4=4\) more lunch boxes.

\(7\) full cartons; \(4\) more lunch boxes are needed for one additional carton.

- Think of a full row as giving one sticker to each child. - Count the completed rows. - How many dots are left in the incomplete row?

1. The array contains \(56\) dots. Each complete row has \(9\) dots, one for each child. 2. Six complete rows give each child \(6\) stickers: \(9\times6=54\). 3. Two dots remain, so \(56\div9=6\) remainder \(2\).

Each child gets \(6\) stickers; \(2\) stickers remain.

- Keep every place of the dividend represented in the quotient. - Ask how many groups of \(4\) are in the middle partial dividend. - A zero quotient digit can be a necessary placeholder.

1. The calculation is \(804\div4\). 2. \(8\div4=2\), so the first quotient digit is \(2\). 3. The next partial dividend is \(0\). The divisor \(4\) fits into \(0\) zero times, so a \(0\) must be written in the tens place of the quotient. 4. Finally, \(4\div4=1\), giving quotient \(201\).

The zero is needed because the tens-place partial dividend is \(0\). The quotient is \(201\).

- Compare the first dividend digit with the divisor. - If the divisor does not fit into the first digit, use the first two digits as the initial partial dividend. - Follow the alignment shown in the written algorithm.

1. The calculation is \(3126\div6\). 2. The first digit \(3\) is smaller than \(6\), so division cannot begin with \(3\) alone. 3. Use the first two digits: \(31\div6=5\) with \(1\) left, so the first quotient digit aligns with the second dividend digit. 4. The later steps are \(12\div6=2\) and \(6\div6=1\), giving quotient \(521\).

Division begins with \(31\) because \(3<6\). The quotient is \(521\).

- Focus on the step whose partial dividend is \(12\). - The quotient digit and subtrahend in one long-division step are linked by multiplication.

1. The second partial dividend is \(12\). 2. Three groups of \(4\) fit in \(12\), so the hidden quotient digit is \(3\). 3. The subtrahend for that step is \(3\times4=12\).

The quotient digit is \(3\), the subtrahend is \(12\), and the connecting fact is \(3\times4=12\).

- Match the hidden subtrahend with the quotient digit directly above its step. - Multiply that quotient digit by the divisor. - Check that the subtraction leaves the remainder needed for the next partial dividend.

1. The calculation is \(1568\div7=224\). 2. In the middle step, the quotient digit is \(2\), so the subtrahend must be \(2\times7=14\). 3. Therefore, the hidden digit in \(1\square\) is \(4\). 4. The remaining steps confirm the quotient \(224\) with no remainder.

The missing digit is \(4\).

- Track the place represented by every dividend digit. - What quotient digit corresponds to a partial dividend smaller than the divisor? - Omitting that zero would change the place values of the later quotient digits.

1. The calculation is \(2835\div7\). 2. \(28\div7=4\), so the hundreds quotient digit is \(4\). 3. The next partial dividend is \(3\), which contains \(0\) groups of \(7\). A \(0\) must be written in the tens place of the quotient. 4. Bringing down \(5\) gives \(35\), and \(35\div7=5\). 5. The quotient is \(405\).

The zero is a necessary tens-place placeholder because \(3<7\). The quotient is \(405\).

- Follow the dividend digits in each panel from left to right. - Compare where the zero partial dividend or too-small partial dividend occurs. - Keep quotient place value aligned with the corresponding dividend place.

1. In a), \(1632\div8=204\). After \(16\div8=2\), the next partial dividend is \(3\), which is smaller than \(8\), so the tens quotient digit is \(0\). 2. In b), \(1680\div8=210\). The final partial dividend is \(0\), so the quotient ends in \(0\). 3. Thus \(204\) has an internal zero, while \(210\) has a zero in the ones place.

a) \(204\) has the internal zero. b) It is needed because the tens-place partial dividend is \(3<8\). c) The quotients are \(204\) and \(210\).

- Compare the displayed remainder with the divisor. - How much is missing from the incomplete final group? - What happens to the quotient when one full extra group is added?

1. The displayed division is \(632\div5=126\) remainder \(2\). 2. The next multiple of \(5\) is \(5-2=3\) higher, so add \(3\). 3. That completes one more group of \(5\), increasing the quotient from \(126\) to \(127\).

a) Add \(3\). b) New quotient: \(127\)

- Look only at the final partial dividend and subtrahend first. - The remainder is what is left after the last subtraction. - Check that the remainder is smaller than \(6\).

1. The calculation is \(745\div6\). 2. The final partial dividend is \(25\), and the final subtrahend is \(24\). 3. \(25-24=1\), so the hidden remainder is \(1\). 4. The complete result is \(745\div6=124\) remainder \(1\).

The remainder is \(1\).

- What remains after the first subtraction? - Look at what is shown after the next digit is brought down. - Work backward from the next subtraction to the missing number.

1. The first subtraction is \(22-20=2\). Bring down the next digit. The next number to divide must begin with \(2\). 2. The next subtraction takes away \(24\). The following step shows \(16\), formed by bringing down the last digit \(6\). So the subtraction before it must leave \(1\). 3. The missing number in the middle step is \(24+1=25\). Its final digit, \(5\), is the missing digit in the original number. 4. Check: \(25-24=1\), then bring down \(6\) to get \(16\), and \(16-16=0\). The completed dividend is \(2256\).

Missing dividend digit: \(5\) Missing number in the written work: \(25\)

- Work out the three divisions independently before comparing. - Keep the remainders from the two separate divisions. - What could account for a difference between the two methods?

1. \((25+11)\div6=36\div6=6\). 2. \(25\div6=4\) remainder \(1\); \(11\div6=1\) remainder \(5\). Adding the whole-number quotients gives \(4+1=5\). 3. The remainders make an extra group of \(6\), because \(1+5=6\). Therefore, adding separate whole-number quotients need not give the quotient of the combined dividend.

1) \(6\). 2) \(4\) remainder \(1\); \(1\) remainder \(5\); quotient sum \(5\). 3) They differ by \(1\) because the two remainders form another full group of \(6\).

- In each step, the subtrahend equals the divisor times the quotient digit for that place. - Use the first quotient digit and first subtrahend together. - Check the recovered divisor against the later subtraction steps.

1. The quotient is \(216\), so the first quotient digit is \(2\). 2. The first subtrahend is \(14\). It must equal \(2\times\text{divisor}\). 3. Since \(2\times7=14\), the hidden divisor is \(7\). 4. The full check is \(1512\div7=216\).

The divisor is \(7\).

- Check each number subtracted against the quotient digit directly above it. - After correcting the first wrong step, continue from the corrected remainder. - Check the result using multiplication and the remainder.

1. The first shown quotient digit is \(3\), so the first subtraction uses \(3\times6=18\), which is correct. 2. The next quotient digit is \(5\), so the next number subtracted should be \(5\times6=30\), not the displayed \(24\). 3. Correcting that step gives \(33-30=3\). Bring down \(4\) to make \(34\). 4. Now \(34\div6=5\) remainder \(4\). The complete division is \(2134\div6=355\) remainder \(4\).

The second subtraction is wrong: subtract \(30\), not \(24\). Correct quotient: \(355\), remainder: \(4\)

- Follow the steps one digit at a time after each subtraction. - Check which dividend digit has not yet been used. - Correct the first misplaced step before finishing the division.

1. The first step is correct: \(32-32=0\). 2. Bring down only the next unused digit. The next number to divide should be \(6\), not \(65\). 3. Divide \(6\) by \(4\): write \(1\) in the quotient, subtract \(4\), and keep a remainder of \(2\). 4. Bring down the last digit \(5\) to make \(25\). Divide \(25\) by \(4\) to get \(6\) with remainder \(1\). 5. The correct result is \(3265\div4=816\) remainder \(1\).

First wrong number: \(65\); it should be \(6\). Quotient: \(816\); remainder: \(1\)

- Work backward from the quotient using multiplication. - Different valid remainders can produce the same whole-number quotient. - Decide how many leftovers are possible before an extra group is made.

1. The dividend is \(5\times84\) plus the remainder. 2. With remainder \(0\), the least dividend is \(5\times84=420\). 3. The greatest remainder when dividing by \(5\) is \(4\), giving \(420+4=424\). 4. A smaller dividend gives a quotient below \(84\); a dividend of \(425\) would make another group of \(5\) and give quotient \(85\).

Smallest: \(420\); greatest: \(424\). All possible starting numbers are \(420,421,422,423,424\). The remainder must be less than \(5\), so only \(0,1,2,3,4\) can be added to \(5\times84\).