Free Math Worksheets

- 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. \(480 \div 6 = 80\).

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

- 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. For part a, \(35\) is the greatest multiple of \(5\) less than \(38\). Since \(35 \div 5 = 7\), the remainder is \(38 - 35 = 3\). 2. For part b, \(42 \div 6 = 7\), and \(47 - 42 = 5\). 3. For part c, \(56 \div 7 = 8\), and \(60 - 56 = 4\). 4. For part d, \(28 \div 4 = 7\), and \(29 - 28 = 1\). 5. For part e, \(81 \div 9 = 9\), and \(85 - 81 = 4\).

a) \(7\) remainder \(3\) b) \(7\) remainder \(5\) c) \(8\) remainder \(4\) d) \(7\) remainder \(1\) e) \(9\) remainder \(4\)

- 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.

- Find the change between consecutive dividends. - Keep the divisor fixed. - Compare consecutive quotients for a second constant change.

1. The dividend increases by \(120\) each step, so the next dividends are \(480\) and \(600\). 2. \(480 \div 6 = 80\) and \(600 \div 6 = 100\). 3. With the divisor fixed at \(6\), increasing the dividend by \(120\) increases the quotient by \(120 \div 6 = 20\).

\(480 \div 6 = 80\) \(600 \div 6 = 100\) The dividend increases by \(120\), and the quotient increases by \(20\).

- Use multiplication as the inverse of division. - Multiply \(90\) by several different divisors. - Check that every dividend is greater than \(200\).

1. Choose different one-digit divisors, such as \(3\), \(4\), and \(5\). 2. Multiply each divisor by \(90\): \(3 \times 90 = 270\), \(4 \times 90 = 360\), and \(5 \times 90 = 450\). 3. Therefore, \(270 \div 3 = 90\), \(360 \div 4 = 90\), and \(450 \div 5 = 90\). Each dividend is greater than \(200\).

One possible set is \(270 \div 3 = 90\), \(360 \div 4 = 90\), and \(450 \div 5 = 90\).

- 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. For \(5\) equal addends, divide by \(5\): \(1{,}000{,}000 \div 5 = 200{,}000\). 2. For \(8\) equal addends, divide by \(8\): \(1{,}000{,}000 \div 8 = 125{,}000\). 3. For \(4\) equal addends, divide by \(4\): \(1{,}000{,}000 \div 4 = 250{,}000\).

a) \(200{,}000\) b) \(125{,}000\) c) \(250{,}000\)

- Equal sharing calls for division. - Divide \(1{,}000{,}000\) by the number of warehouses. - As the number of warehouses increases, each share becomes smaller.

1. For \(2\) warehouses, \(1{,}000{,}000 \div 2 = 500{,}000\). 2. For \(4\) warehouses, \(1{,}000{,}000 \div 4 = 250{,}000\). 3. For \(5\) warehouses, \(1{,}000{,}000 \div 5 = 200{,}000\). 4. For \(8\) warehouses, \(1{,}000{,}000 \div 8 = 125{,}000\).

a) \(500{,}000\) screws b) \(250{,}000\) screws c) \(200{,}000\) screws d) \(125{,}000\) screws

- Divide the number at the top of a column by the given factor. - Dividing by \(4\) can be done by halving twice. - Look for the factor-of-\(10\) pattern across each row.

1. For the \(10{,}000\) column, divide by each given factor: \(10{,}000 \div 2 = 5000\), \(10{,}000 \div 4 = 2500\), \(10{,}000 \div 5 = 2000\), and \(10{,}000 \div 8 = 1250\). 2. For the \(100{,}000\) column, the missing factors are \(50{,}000\), \(25{,}000\), \(20{,}000\), and \(12{,}500\). 3. For the \(1{,}000{,}000\) column, the missing factors are \(500{,}000\), \(250{,}000\), \(200{,}000\), and \(125{,}000\).

<table> <tr> <th>Given factor</th> <th>\(10{,}000\)</th> <th>\(100{,}000\)</th> <th>\(1{,}000{,}000\)</th> </tr> <tr> <td>\(2 \times \square\)</td> <td>\(5000\)</td><td>\(50{,}000\)</td><td>\(500{,}000\)</td> </tr> <tr> <td>\(4 \times \square\)</td> <td>\(2500\)</td><td>\(25{,}000\)</td><td>\(250{,}000\)</td> </tr> <tr> <td>\(5 \times \square\)</td> <td>\(2000\)</td><td>\(20{,}000\)</td><td>\(200{,}000\)</td> </tr> <tr> <td>\(8 \times \square\)</td> <td>\(1250\)</td><td>\(12{,}500\)</td><td>\(125{,}000\)</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) \(1307\)

- What number near \(26\) is easy to divide by \(5\)? - How many cents are in \(\$1.00\)? - Convert the total price to cents before dividing.

1. Round \(26\,\text{oz}\) to \(25\,\text{oz}\), which is easy to divide by \(5\). Then \(25\,\text{oz} \div 5 = 5\,\text{oz}\), so one peach weighs about \(5\,\text{oz}\). 2. Convert the total cost to cents: \(\$2.45 = 245\) cents. Then \(245 \div 5 = 49\), so one peach costs \(49\) cents.

a) One peach weighs about \(5\,\text{oz}\). b) One peach costs \(49\) cents.

- What whole number near \(31.8\) is easy to divide by \(4\)? - Convert the dollar amount to cents before dividing. - Break \(1160\) into parts that are easy to divide by \(4\).

1. Round \(31.8\,\text{lb}\) to \(32\,\text{lb}\). Then \(32\,\text{lb} \div 4 = 8\,\text{lb}\), so one package weighs about \(8\,\text{lb}\). 2. Convert the shipping cost to cents: \(\$11.60 = 1160\) cents. Then \(1160 \div 4 = 290\), so shipping costs \(290\) cents per package.

a) One package weighs about \(8\,\text{lb}\). b) Shipping costs \(290\) cents, or \(\$2.90\), per package.

- 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.

- 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) \(100\)

- 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.

- 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. For part a, \(8 \times 8 = 64\), and \(67 - 64 = 3\), so the result is \(8\) remainder \(3\). 2. For part b, \(9 \times 9 = 81\), and \(84 - 81 = 3\), so the result is \(9\) remainder \(3\). 3. For part c, \(7 \times 7 = 49\), and \(55 - 49 = 6\), so the result is \(7\) remainder \(6\). 4. For part d, \(9 \times 10 = 90\), and \(93 - 90 = 3\), so the result is \(9\) remainder \(3\). 5. For part e, \(8 \times 9 = 72\), and \(76 - 72 = 4\), so the result is \(8\) remainder \(4\).

a) \(8\) remainder \(3\) b) \(9\) remainder \(3\) c) \(7\) remainder \(6\) d) \(9\) remainder \(3\) 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) \(13\)

- 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 leading digit of the dividend with the divisor. - If the leading digit is too small, division begins with the first two digits. - Compare that starting point with the total number of dividend digits.

1. For a), the leading digit \(8\) is at least the divisor \(5\), so the quotient has the same number of digits as the dividend: \(5\) digits. 2. For b), the leading digit \(2\) is less than \(6\), so division begins with \(24\). The quotient has one fewer digit than the five-digit dividend: \(4\) digits. 3. For c), the leading digit \(3\) is less than \(9\), so division begins with \(36\). The quotient has one fewer digit than the six-digit dividend: \(5\) digits.

a) \(5\) digits b) \(4\) digits c) \(5\) digits

- 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.

- 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.

- Use multiplication to find a missing dividend. - Divide the dividend by the quotient to find a missing divisor. - Check every equation.

1. In a), the missing divisor is \(120 \div 40 = 3\). 2. In b), the missing dividend is \(40 \times 5 = 200\). 3. In c), \(280 \div 7 = 40\). 4. In d), \(360 \div 9 = 40\). 5. In e), the missing dividend is \(40 \times 2 = 80\).

a) \(3\) b) \(200\) c) \(40\) d) \(40\) 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: \(8 \times 9 + 2 = 74\)

- 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 \(5\).

- 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 \(161\) 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 \(3\). - Use place value when dividing by \(10\).

1. \(840 \div 2 = 420\), so \(840 \div 4 = 210\), which is half of \(420\). 2. \(840 \div 3 = 280\), so \(840 \div 6 = 140\), which is half of \(280\). 3. \(840 \div 7 = 120\), and \(840 \div 10 = 84\).

a) \(420\) and \(210\) b) \(280\) and \(140\) c) \(120\) and \(84\)

- Find both quotients in each comparison. - Write the two values before choosing a comparison symbol. - Use related multiplication facts to divide.

1. \(240 \div 4 = 60\) and \(180 \div 3 = 60\), so the quotients are equal. 2. \(560 \div 7 = 80\) and \(540 \div 6 = 90\), so \(80 < 90\). 3. \(420 \div 6 = 70\) and \(350 \div 5 = 70\), so the quotients are equal. 4. \(320 \div 8 = 40\) and \(270 \div 9 = 30\), so \(40 > 30\).

1) \(=\) 2) \(<\) 3) \(=\) 4) \(>\)

- 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 \(5\). - Use place value to divide by \(10\). - Since \(10\) is twice \(5\), compare the two quotients in each row.

1. For \(200\): \(200 \div 5 = 40\) and \(200 \div 10 = 20\). 2. For \(350\): \(350 \div 5 = 70\) and \(350 \div 10 = 35\). 3. For \(600\): \(600 \div 5 = 120\) and \(600 \div 10 = 60\). 4. For \(850\): \(850 \div 5 = 170\) and \(850 \div 10 = 85\).

<table> <tr><th>Number</th><th>Divided by \(5\)</th><th>Divided by \(10\)</th></tr> <tr><td>\(200\)</td><td>\(40\)</td><td>\(20\)</td></tr> <tr><td>\(350\)</td><td>\(70\)</td><td>\(35\)</td></tr> <tr><td>\(600\)</td><td>\(120\)</td><td>\(60\)</td></tr> <tr><td>\(850\)</td><td>\(170\)</td><td>\(85\)</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.

- Convert the total price to cents. - Divide the number of cents equally among \(3\) children. - Convert the result back to dollars.

1. Write the total price in cents: \(\$7.50 = 750\text{ cents}\). 2. Divide the total by \(3\): \(750 \div 3\). 3. Break \(750\) into compatible numbers: \(600 \div 3 = 200\) and \(150 \div 3 = 50\). Then add: \(200 + 50 = 250\) cents. 4. Convert back to dollars: \(250\text{ cents} = \$2.50\).

Each child pays \(\$2.50\).

- 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.

- Find the greatest multiple of \(6\) that does not exceed \(35\). - The remainder must be less than \(6\).

1. Five full rows contain \(5 \times 6 = 30\) dots. 2. The number of dots left over is \(35 - 30 = 5\). 3. Therefore, \(35 \div 6 = 5\) remainder \(5\).

\(35 \div 6 = 5\) remainder \(5\). There are \(5\) full rows and \(5\) dots left over.

- Mentally circle groups of \(5\) dots. - After making as many complete teams as possible, count the dots that remain.

1. Eight teams of \(5\) include \(8 \times 5 = 40\) students. 2. Subtract to find the remainder: \(42 - 40 = 2\). 3. Therefore, \(42 \div 5 = 8\) remainder \(2\).

There are \(8\) complete teams, with \(2\) students remaining. In symbols, \(42 \div 5 = 8\) remainder \(2\).

- Make groups of \(3\) in the array. - Check your quotient with multiplication.

1. Separate the \(42\) dots into groups of \(3\). 2. There are \(14\) groups, so \(42 \div 3 = 14\).

The medals are enough for \(14\) teams because \(42 \div 3 = 14\).

- Count the complete rows of \(9\). - For the check, add the remainder to the product of the quotient and divisor.

1. There are \(8\) complete rows of \(9\), so the full envelopes hold \(8\times 9=72\) tickets. 2. The remaining partial row contains \(5\) tickets. 3. Therefore, \(77\div 9=8\text{ R }5\). 4. Check: \(8\times 9+5=77\).

\(77\div 9=8\text{ R }5\). Check: \(8\times 9+5=77\).

- 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.

- Use multiples of \(9\) near each dividend. - Think about how many objects can remain without making another full group of \(9\). - A valid remainder must be less than the divisor.

1. The results are \(79 \div 9 = 8\) remainder \(7\), \(80 \div 9 = 8\) remainder \(8\), \(81 \div 9 = 9\) remainder \(0\), and \(82 \div 9 = 9\) remainder \(1\). 2. There are \(9\) possible remainders: \(0\), \(1\), \(2\), \(3\), \(4\), \(5\), \(6\), \(7\), and \(8\). 3. A remainder of \(10\) is greater than the divisor \(9\). One more group of \(9\) could be removed, leaving a remainder of \(1\).

a) \(8\) remainder \(7\); \(8\) remainder \(8\); \(9\) remainder \(0\); \(9\) remainder \(1\) b) The possible remainders are \(0\) through \(8\). c) A remainder must be less than the divisor. From \(10\), another group of \(9\) can be removed.

- Find the quotient and remainder in parts a and b. - In part c, look for the next number after \(23\) that leaves remainder \(4\) when divided by \(5\). - A remainder must be less than the divisor.

1. For part a, \(23 \div 4 = 5\) remainder \(3\). 2. For part b, \(23 \div 5 = 4\) remainder \(3\). 3. One more than the remainder in part b is \(4\). The least number greater than \(23\) that has remainder \(4\) when divided by \(5\) is \(24\), because \(24 \div 5 = 4\) remainder \(4\). 4. The greatest possible remainder when dividing by \(5\) is \(4\), because a remainder must be less than the divisor.

a) \(3\) notebooks b) \(3\) notebooks c) \(24\) notebooks d) The greatest possible remainder is \(4\).

- 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 quotient by its divisor. - Compare each product with the original dividend. - When correcting part c, remember that a zero may appear in the quotient.

1. Check part a: \(208\times 3=624\), so it is correct. 2. Check part b: \(149\times 5=745\), so it is correct. 3. Check part c: \(14\times 9=126\), and \(126\neq 936\), so it is incorrect. 4. The correct quotient is \(936\div 9=104\).

c) is incorrect. The check gives \(14\times 9=126\). The correct equation is \(936\div 9=104\).

- Divide the total mass equally among the \(9\) plums. - Think of a nearby number to \(400\) that is divisible by \(9\). - A quotient with a remainder can be used for the estimate.

1. Divide the total mass by the number of plums: \(400\,\text{g} \div 9\). 2. \(400 \div 9 = 44\) remainder \(4\), so each plum has a mass of about \(44\,\text{g}\).

Each plum has a mass of 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: \(100 + 14 = 114\).

\(114\)

- Find multiples of \(15\) greater than \(100\). - Use multiplication to create each dividend. - Check that every dividend is less than \(200\).

1. Find multiples of \(15\) between \(100\) and \(200\). 2. For example, \(7 \times 15 = 105\), \(8 \times 15 = 120\), and \(9 \times 15 = 135\). 3. Therefore, \(105 \div 7 = 15\), \(120 \div 8 = 15\), and \(135 \div 9 = 15\).

One possible set is \(105 \div 7 = 15\), \(120 \div 8 = 15\), and \(135 \div 9 = 15\).

- 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) \(80\) (mental math) b) \(211\) (partial quotients) c) \(224\) (standard algorithm)

- 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) \(500\) (mental math) b) \(511\) (partial quotients) c) \(703\) (standard algorithm) d) \(604\) (partial quotients)

- 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 multiplying it by the divisor.

1. Divide the award equally among the four schools: \(\$12{,}448 \div 4 = \$3112\). 2. Divide one school’s share equally among its eight classes: \(\$3112 \div 8 = \$389\). 3. Therefore, each class receives \(\$389\).

Each class receives \(\$389\).

- 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.

- Use nearby multiples of \(7\). - The remainder is the distance from the dividend to the greatest multiple of \(7\) below it. - A remainder must be less than \(7\).

1. \(560 \div 7=80\). 2. Since \(566=560+6\), \(566 \div 7=80\text{ R }6\). 3. The greatest multiple of \(7\) not exceeding \(558\) is \(553=7 \times 79\). Therefore, \(558 \div 7=79\text{ R }5\).

a) \(80\) b) \(80\text{ R }6\) c) \(79\text{ R }5\)

- 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 \(1\).

- One meter equals \(100\) centimeters. - Convert the measurement before dividing. - Multiply the quotient by \(8\) to check the division.

1. Convert meters to centimeters: \(145.60\,\text{m} \times 100=14{,}560\,\text{cm}\). 2. Divide: \(14{,}560\,\text{cm} \div 8=1820\,\text{cm}\). 3. Check: \(1820\,\text{cm} \times 8=14{,}560\,\text{cm}\).

The result is \(1820\,\text{cm}\). Check: \(1820\,\text{cm} \times 8=14{,}560\,\text{cm}\).

- 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) Mental math: \(210\) b) Standard algorithm: \(136\)

- 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.

- 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 a result by multiplying the quotient by the divisor and adding the remainder. - A valid remainder must be less than the divisor. - When a remainder is too large, see whether another full group of the divisor can be made.

1. For part a, \(7 \times 5 + 3 = 35 + 3 = 38\), and the remainder \(3\) is less than the divisor \(5\). The statement is correct. 2. For part b, \(6 \times 6 + 10 = 36 + 10 = 46\), but the remainder \(10\) is greater than the divisor \(6\). Another group of \(6\) can be made, so the correct result is \(46 \div 6 = 7\) remainder \(4\). 3. For part c, \(7 \times 8 + 3 = 56 + 3 = 59\), and the remainder \(3\) is less than the divisor \(8\). The statement is correct.

a) Correct. b) Incorrect. A remainder must be less than the divisor. The correct result is \(46 \div 6 = 7\) remainder \(4\). 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) \(3\)

- 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.

- 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) \(=\)

- 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 \(4\).

- The greatest valid remainder is \(1\) less than the divisor. - Multiply the quotient by the divisor after finding the remainder. - Add the remainder to that product to find the dividend.

1. A remainder must be less than the divisor, so the greatest possible remainder is always \(1\) less than the divisor. 2. For part a, the greatest remainder is \(3\). The dividend is \(8 \times 4 + 3 = 32 + 3 = 35\). 3. For part b, the greatest remainder is \(6\). The dividend is \(5 \times 7 + 6 = 35 + 6 = 41\). 4. For part c, the greatest remainder is \(8\). The dividend is \(6 \times 9 + 8 = 54 + 8 = 62\). 5. For part d, the greatest remainder is \(5\). The dividend is \(7 \times 6 + 5 = 42 + 5 = 47\).

a) \(35 \div 4 = 8\) remainder \(3\) b) \(41 \div 7 = 5\) remainder \(6\) c) \(62 \div 9 = 6\) remainder \(8\) d) \(47 \div 6 = 7\) 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 \(1\).

- Calculate every quotient first. - Compare the quotient values by place value. - A quotient is the result of a division.

1. \(585 \div 5=117\). 2. \(756 \div 6=126\). 3. \(945 \div 7=135\). 4. \(812 \div 4=203\). 5. Therefore, \(117<126<135<203\).

\(585 \div 5\), \(756 \div 6\), \(945 \div 7\), \(812 \div 4\)

- Use multiplication to reverse division. - To reconstruct a dividend, multiply the divisor and quotient, then add the remainder. - To find a remainder, subtract the greatest represented multiple from the dividend.

1. For a), reconstruct the dividend: \(123 \times 4+2=492+2=494\). 2. For b), \(105 \times 7=735\), so the remainder is \(739-735=4\). 3. For c), \(846 \div 141=6\), and \(141 \times 6=846\).

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).

- Match the divisor in each method to the divisor in the problem. - Check that the two partial dividends add to the original dividend. - Add the partial quotients.

1. Problem A matches Method 2 because \(70 + 42 = 112\) and both parts are divided by \(7\). The quotient is \(10 + 6 = 16\). 2. Problem B matches Method 1 because \(60 + 48 = 108\) and both parts are divided by \(6\). The quotient is \(10 + 8 = 18\).

A matches 2: \(112 \div 7 = 16\) B matches 1: \(108 \div 6 = 18\)

- 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 \(9\).

- 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.

- 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 on the trip. - For parts a) and b), divide and identify each remainder. - Compare the two remainders for part c).

1. Find the total number of children: \(28 + 25 = 53\). 2. For part a), divide by six: \(53 \div 6 = 8\) remainder \(5\). 3. For part b), divide by five: \(53 \div 5 = 10\) remainder \(3\). 4. Since \(3 < 5\), groups of five leave fewer children left over.

a) \(5\) children are left over. b) \(3\) children are left over. c) Groups of \(5\) leave fewer children left over.

- 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 \(3\).

- List multiples of \(7\) near the interval from \(30\) through \(50\). - Add \(4\) to each multiple of \(7\). - Keep only the results that are from \(30\) through \(50\).

1. Numbers with remainder \(4\) when divided by \(7\) have the form \(7 \times q + 4\). 2. Using \(q = 4\) gives \(7 \times 4 + 4 = 28 + 4 = 32\). 3. Using \(q = 5\) gives \(7 \times 5 + 4 = 35 + 4 = 39\). 4. Using \(q = 6\) gives \(7 \times 6 + 4 = 42 + 4 = 46\). 5. The next value is \(7 \times 7 + 4 = 53\), which is greater than \(50\). The previous value is \(7 \times 3 + 4 = 25\), which is less than \(30\). Therefore, the complete list is \(32\), \(39\), and \(46\).

\(32 \div 7 = 4\) remainder \(4\) \(39 \div 7 = 5\) remainder \(4\) \(46 \div 7 = 6\) 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 \(7\).

- Find a large multiple of \(9\) close to \(855\). - Divide both parts by \(9\). - Add the partial quotients and explain why the chosen parts are useful.

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 \(9\).

- The greatest possible remainder is \(1\) less than the divisor. - Use the greatest listed remainder to find the divisor. - For part b, add \(3\) to a multiple of the divisor.

1. The greatest possible remainder is always \(1\) less than the divisor. Since the greatest possible remainder is \(5\), the divisor is \(6\). 2. For part b, choose a multiple of \(6\) and add \(3\). For example, \(2 \times 6 + 3 = 15\), so \(15 \div 6 = 2\) remainder \(3\).

a) The divisor is \(6\). b) One possible example is \(15 \div 6 = 2\) 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 the student’s quotient with multiplication. - Bring down and divide every digit of the dividend. - Write a zero in the quotient when a partial dividend is smaller than the divisor.

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. The student 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) The student 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: \(1017\times 8=8136\).

- 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 \(\$85\).

- 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.

- 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. - Divide the dividend by the quotient to find a missing divisor. - Break a dividend into convenient parts when finding a quotient.

1. In a), the missing dividend is \(120 \times 4 = 480\). 2. In b), the missing divisor is \(560 \div 70 = 8\). 3. In c), \(840 \div 2 = 420\). 4. In d), the missing dividend is \(210 \times 3 = 630\). 5. In e), use partial quotients: \(500 \div 5 = 100\) and \(150 \div 5 = 30\), so \(650 \div 5 = 130\).

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

- 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 division. - Add the remainder after multiplying the quotient by the divisor. - Check that \(61 \div 7\) has quotient \(8\) and remainder \(5\).

1. Multiply the quotient by the divisor: \(8 \times 7 = 56\). 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.

- 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 \(\$28\).

- 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.

The quotients are \(102\), \(120\), \(201\), and \(210\). The order is \(306 \div 3\), \(360 \div 3\), \(603 \div 3\), \(630 \div 3\).

- 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 \(104 \div 8 = 13\).

- Estimate which quotient in each pair should be larger. - Find both quotients in each pair. - Write the two results side by side. - Compare the ones and tens places carefully.

1. For part a, \(612 \div 6 = 102\) and \(618 \div 6 = 103\), so \(618 \div 6\) is greater. 2. For part b, \(832 \div 8 = 104\) and \(832 \div 4 = 208\), so \(832 \div 4\) is greater. 3. For part c, \(918 \div 9 = 102\) and \(412 \div 4 = 103\), so \(412 \div 4\) is greater.

a) \(618 \div 6\) b) \(832 \div 4\) c) \(412 \div 4\)

- 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.

- Check each result by multiplying the quotient by the divisor and adding the remainder. - A valid remainder must be less than the divisor. - Make sure each check equals the original dividend.

1. Lucas: \(8 \times 9 + 4 = 72 + 4 = 76\), and \(4 < 9\). His result is correct. 2. Mia: \(7 \times 9 + 13 = 76\), but \(13 > 9\). The remainder is not less than the divisor, so another group of \(9\) must be included in the quotient. 3. Sophie: \(8 \times 9 + 3 = 75\), not \(76\). Her result does not reconstruct the dividend.

Lucas is correct. Mia is incorrect because the remainder \(13\) is greater than the divisor \(9\). Sophie is incorrect because \(8 \times 9 + 3 = 75\), 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 an equation must have the same value. - Use multiplication or division as the inverse operation to find the missing number.

1. For a), \(324 \div 4=81\). The missing dividend must satisfy \(\Box \div 9=81\), so \(81 \times 9=729\). 2. For b), \(735 \div 7=105\). The missing divisor must satisfy \(525 \div \Box=105\), so \(525 \div 105=5\).

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 \(3\).

- Count only complete groups of \(4\). - Compare the number left over with the bag size of \(4\).

1. Seven full bags use \(7 \times 4 = 28\) marbles. 2. The number left over is \(29 - 28 = 1\). 3. Another full bag would need \(4\) marbles, so \(1\) marble is not enough. Therefore, \(29 \div 4 = 7\) remainder \(1\).

There are \(7\) full bags and \(1\) marble left over. There are not enough marbles for another full bag.

- Calculate \(6 \times 7\). - Subtract that product from \(45\).

1. Six full groups of \(7\) contain \(6 \times 7 = 42\) dots. 2. The number left over is \(45 - 42 = 3\), not \(2\). 3. The correct equation is \(45 \div 7 = 6\) remainder \(3\).

The correct equation is \(45 \div 7 = 6\) remainder \(3\). Noah made the remainder \(1\) too small.

- Find the greatest multiple of \(8\) that does not exceed \(50\). - Compare any proposed remainder with the divisor.

1. Six full rows contain \(6 \times 8 = 48\) dots. 2. The number left over is \(50 - 48 = 2\), so \(50 \div 8 = 6\) remainder \(2\). 3. A remainder of \(9\) is impossible because a remainder must be less than the divisor \(8\). A remainder of \(9\) would contain another full group of \(8\).

\(50 \div 8 = 6\) remainder \(2\). A remainder of \(9\) is impossible because it is greater than the divisor \(8\).

- First subtract the damaged eggs. - Then divide the undamaged eggs into groups of six.

1. Find the number of undamaged eggs: \(50 - 7 = 43\). 2. Divide the undamaged eggs into groups of six: \(43 \div 6 = 7\) remainder \(1\). 3. Therefore, \(7\) full cartons can be filled, and \(1\) egg is left over.

\(7\) full cartons can be filled, and \(1\) egg is left over.

- Find the greatest multiple of \(8\) that does not exceed \(60\). - Compare the remainder with \(8\).

1. Seven full cartons hold \(7 \times 8 = 56\) lunch boxes. 2. The number left over is \(60 - 56 = 4\). 3. One carton holds \(8\), so the number needed to fill another carton is \(8 - 4 = 4\).

There are \(7\) full cartons. Four lunch boxes are left over, and \(4\) more are needed to fill another carton.

- Compare \(9 \times 6\) and \(9 \times 7\) with \(56\). - Use the greatest product that does not exceed \(56\).

1. Giving each child \(6\) stickers uses \(9 \times 6 = 54\) stickers. 2. The number left over is \(56 - 54 = 2\). 3. Therefore, \(56 \div 9 = 6\) remainder \(2\).

Each child receives \(6\) stickers, and \(2\) stickers are left over.

- Convert all three measurements to centimeters first. - Divide each converted value by \(9\). - Compare how the original measurements and the quotients are scaled.

1. Convert to centimeters: \(105.30\,\text{m}=10{,}530\,\text{cm}\), \(210.60\,\text{m}=21{,}060\,\text{cm}\), and \(315.90\,\text{m}=31{,}590\,\text{cm}\). 2. Divide: \(10{,}530\,\text{cm} \div 9=1170\,\text{cm}\), \(21{,}060\,\text{cm} \div 9=2340\,\text{cm}\), and \(31{,}590\,\text{cm} \div 9=3510\,\text{cm}\). 3. The second quotient is twice the first, and the third quotient is three times the first, matching the scale factors of the original lengths.

1) \(1170\,\text{cm}\) 2) \(2340\,\text{cm}\) 3) \(3510\,\text{cm}\) The second quotient is twice the first, and the third quotient is three times the first.

- Record both the quotient and remainder for each separate division. - Add the remainders as well as comparing the quotients. - Check whether the combined remainders make another full group of \(6\).

1. \((25 + 11) \div 6 = 36 \div 6 = 6\). 2. \(25 \div 6 = 4\) remainder \(1\), and \(11 \div 6 = 1\) remainder \(5\). The whole-number quotients add to \(4 + 1 = 5\). 3. The results differ because the remainders combine to make another group of \(6\). Therefore, adding only the separate whole-number quotients does not always give the quotient of the combined dividend.

The combined quotient is \(6\), but the separate whole-number quotients add to \(5\). The remainders \(1\) and \(5\) form one additional group of \(6\).

- For each estimate, choose a nearby dividend that is easy to divide by the divisor. - In the standard algorithm, write a zero in the quotient when the current place value is smaller than the divisor. - Check each result by multiplying the quotient by the divisor.

1. Estimate: \(4800 \div 8 = 600\). Using the standard algorithm, \(4656 \div 8 = 582\). 2. Estimate: \(420{,}000 \div 6 = 70{,}000\). Using the standard algorithm, \(438{,}204 \div 6 = 73{,}034\).

1) Estimate: \(600\); exact quotient: \(582\) 2) Estimate: \(70{,}000\); exact quotient: \(73{,}034\)