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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

- 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 \(8\)

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

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

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

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

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

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

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

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

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

- 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 \(3\)

- 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: \(8 \times 9 + 4 = 76\).

- 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 \(3\).

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

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

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

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

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

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

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

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

The other factor is \(40\).

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

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

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

- 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 \(3\).

- 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) \(280\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

- 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 \(4\).

- 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: \(2400\div 2=1200\) \(3600\div 3=1200\) \(4800\div 4=1200\) \(6000\div 5=1200\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

- 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.

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

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

\(2\) children help the referee.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(1\) apple is left over.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

- 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 \(2\).

- 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 \(1345\div 5=269\).

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

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

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

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

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

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

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

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

The schools need \(50\) museum guides.

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

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

\(61\)

- 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 \(3\).

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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