← 4-1 · A number divides by its factor pieces · Decompose a Number into Parts and Factors

A number divides by its factor pieces · 10 practice problems

4.OA.B.44.NBT.B.6

Generated variants — 10

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 answer: The two boxes are 3 and 3 (so 1476 = 41 x 2 x 2 x 3 x 3).

The number 1476 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$1476 = 41 \times 2 \times 2 \times \square \times \square$

Show solution

Understand

Write 1476 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 1476 = 41 x 2 x 2 x box x box, find the two missing numbers.

Givens

The number is 1476
It is already partly factored as 41 x 2 x 2 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 1476
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 1476 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 1476 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 1476 by the factors already shown. 1476 / 41 = 36, then 36 / 2 = 18, then 18 / 2 = 9. So the two boxes must multiply to 9.

1476 \div 41 \div 2 \div 2 = 9

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 9 into the smallest whole numbers greater than 1. 9 = 3 x 3, and both 3 and 3 are prime, so they cannot be broken down further.

9 = 3 \times 3

9 has only one factor pair other than 1 and 9, namely 3 and 3, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 3 and 3, giving the full prime factorization 1476 = 41 x 2 x 2 x 3 x 3.

1476 = 41 \times 2 \times 2 \times 3 \times 3

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 3 and 3 (so 1476 = 41 x 2 x 2 x 3 x 3).

Review

Multiply back: 41 x 2 x 2 x 3 x 3 = 1476, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 1476 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 9 into the prime factor pair 3 and 3 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 1476 by the given factors to find the leftover 9.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!

Variant 2 answer: The two boxes are 5 and 7 (so 2660 = 19 x 2 x 2 x 5 x 7).

The number 2660 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$2660 = 19 \times 2 \times 2 \times \square \times \square$

Show solution

Understand

Write 2660 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 2660 = 19 x 2 x 2 x box x box, find the two missing numbers.

Givens

The number is 2660
It is already partly factored as 19 x 2 x 2 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 2660
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 2660 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 2660 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 2660 by the factors already shown. 2660 / 19 = 140, then 140 / 2 = 70, then 70 / 2 = 35. So the two boxes must multiply to 35.

2660 \div 19 \div 2 \div 2 = 35

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 35 into the smallest whole numbers greater than 1. 35 = 5 x 7, and both 5 and 7 are prime, so they cannot be broken down further.

35 = 5 \times 7

35 has only one factor pair other than 1 and 35, namely 5 and 7, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 5 and 7, giving the full prime factorization 2660 = 19 x 2 x 2 x 5 x 7.

2660 = 19 \times 2 \times 2 \times 5 \times 7

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 5 and 7 (so 2660 = 19 x 2 x 2 x 5 x 7).

Review

Multiply back: 19 x 2 x 2 x 5 x 7 = 2660, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 2660 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 35 into the prime factor pair 5 and 7 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 2660 by the given factors to find the leftover 35.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!

Variant 3 answer: The two boxes are 3 and 5 (so 1380 = 23 x 2 x 2 x 3 x 5).

The number 1380 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$1380 = 23 \times 2 \times 2 \times \square \times \square$

Show solution

Understand

Write 1380 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 1380 = 23 x 2 x 2 x box x box, find the two missing numbers.

Givens

The number is 1380
It is already partly factored as 23 x 2 x 2 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 1380
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 1380 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 1380 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 1380 by the factors already shown. 1380 / 23 = 60, then 60 / 2 = 30, then 30 / 2 = 15. So the two boxes must multiply to 15.

1380 \div 23 \div 2 \div 2 = 15

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 15 into the smallest whole numbers greater than 1. 15 = 3 x 5, and both 3 and 5 are prime, so they cannot be broken down further.

15 = 3 \times 5

15 has only one factor pair other than 1 and 15, namely 3 and 5, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 3 and 5, giving the full prime factorization 1380 = 23 x 2 x 2 x 3 x 5.

1380 = 23 \times 2 \times 2 \times 3 \times 5

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 3 and 5 (so 1380 = 23 x 2 x 2 x 3 x 5).

Review

Multiply back: 23 x 2 x 2 x 3 x 5 = 1380, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 1380 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 15 into the prime factor pair 3 and 5 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 1380 by the given factors to find the leftover 15.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!

Variant 4 answer: The two boxes are 2 and 5 (so 1720 = 43 x 2 x 2 x 2 x 5).

The number 1720 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$1720 = 43 \times 2 \times 2 \times \square \times \square$

Show solution

Understand

Write 1720 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 1720 = 43 x 2 x 2 x box x box, find the two missing numbers.

Givens

The number is 1720
It is already partly factored as 43 x 2 x 2 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 1720
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 1720 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 1720 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 1720 by the factors already shown. 1720 / 43 = 40, then 40 / 2 = 20, then 20 / 2 = 10. So the two boxes must multiply to 10.

1720 \div 43 \div 2 \div 2 = 10

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 10 into the smallest whole numbers greater than 1. 10 = 2 x 5, and both 2 and 5 are prime, so they cannot be broken down further.

10 = 2 \times 5

10 has only one factor pair other than 1 and 10, namely 2 and 5, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 2 and 5, giving the full prime factorization 1720 = 43 x 2 x 2 x 2 x 5.

1720 = 43 \times 2 \times 2 \times 2 \times 5

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 2 and 5 (so 1720 = 43 x 2 x 2 x 2 x 5).

Review

Multiply back: 43 x 2 x 2 x 2 x 5 = 1720, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 1720 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 10 into the prime factor pair 2 and 5 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 1720 by the given factors to find the leftover 10.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!

Variant 5 answer: The two boxes are 3 and 7 (so 1386 = 11 x 2 x 3 x 3 x 7).

The number 1386 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$1386 = 11 \times 2 \times 3 \times \square \times \square$

Show solution

Understand

Write 1386 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 1386 = 11 x 2 x 3 x box x box, find the two missing numbers.

Givens

The number is 1386
It is already partly factored as 11 x 2 x 3 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 1386
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 1386 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 1386 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 1386 by the factors already shown. 1386 / 11 = 126, then 126 / 2 = 63, then 63 / 3 = 21. So the two boxes must multiply to 21.

1386 \div 11 \div 2 \div 3 = 21

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 21 into the smallest whole numbers greater than 1. 21 = 3 x 7, and both 3 and 7 are prime, so they cannot be broken down further.

21 = 3 \times 7

21 has only one factor pair other than 1 and 21, namely 3 and 7, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 3 and 7, giving the full prime factorization 1386 = 11 x 2 x 3 x 3 x 7.

1386 = 11 \times 2 \times 3 \times 3 \times 7

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 3 and 7 (so 1386 = 11 x 2 x 3 x 3 x 7).

Review

Multiply back: 11 x 2 x 3 x 3 x 7 = 1386, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 1386 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 21 into the prime factor pair 3 and 7 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 1386 by the given factors to find the leftover 21.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!

Variant 6 answer: The two boxes are 2 and 11 (so 1144 = 13 x 2 x 2 x 2 x 11).

The number 1144 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$1144 = 13 \times 2 \times 2 \times \square \times \square$

Show solution

Understand

Write 1144 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 1144 = 13 x 2 x 2 x box x box, find the two missing numbers.

Givens

The number is 1144
It is already partly factored as 13 x 2 x 2 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 1144
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 1144 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 1144 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 1144 by the factors already shown. 1144 / 13 = 88, then 88 / 2 = 44, then 44 / 2 = 22. So the two boxes must multiply to 22.

1144 \div 13 \div 2 \div 2 = 22

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 22 into the smallest whole numbers greater than 1. 22 = 2 x 11, and both 2 and 11 are prime, so they cannot be broken down further.

22 = 2 \times 11

22 has only one factor pair other than 1 and 22, namely 2 and 11, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 2 and 11, giving the full prime factorization 1144 = 13 x 2 x 2 x 2 x 11.

1144 = 13 \times 2 \times 2 \times 2 \times 11

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 2 and 11 (so 1144 = 13 x 2 x 2 x 2 x 11).

Review

Multiply back: 13 x 2 x 2 x 2 x 11 = 1144, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 1144 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 22 into the prime factor pair 2 and 11 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 1144 by the given factors to find the leftover 22.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!

Variant 7 answer: The two boxes are 2 and 3 (so 888 = 37 x 2 x 2 x 2 x 3).

The number 888 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$888 = 37 \times 2 \times 2 \times \square \times \square$

Show solution

Understand

Write 888 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 888 = 37 x 2 x 2 x box x box, find the two missing numbers.

Givens

The number is 888
It is already partly factored as 37 x 2 x 2 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 888
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 888 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 888 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 888 by the factors already shown. 888 / 37 = 24, then 24 / 2 = 12, then 12 / 2 = 6. So the two boxes must multiply to 6.

888 \div 37 \div 2 \div 2 = 6

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 6 into the smallest whole numbers greater than 1. 6 = 2 x 3, and both 2 and 3 are prime, so they cannot be broken down further.

6 = 2 \times 3

6 has only one factor pair other than 1 and 6, namely 2 and 3, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 2 and 3, giving the full prime factorization 888 = 37 x 2 x 2 x 2 x 3.

888 = 37 \times 2 \times 2 \times 2 \times 3

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 2 and 3 (so 888 = 37 x 2 x 2 x 2 x 3).

Review

Multiply back: 37 x 2 x 2 x 2 x 3 = 888, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 888 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 6 into the prime factor pair 2 and 3 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 888 by the given factors to find the leftover 6.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!

Variant 8 answer: The two boxes are 2 and 5 (so 1240 = 31 x 2 x 2 x 2 x 5).

The number 1240 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$1240 = 31 \times 2 \times 2 \times \square \times \square$

Show solution

Understand

Write 1240 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 1240 = 31 x 2 x 2 x box x box, find the two missing numbers.

Givens

The number is 1240
It is already partly factored as 31 x 2 x 2 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 1240
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 1240 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 1240 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 1240 by the factors already shown. 1240 / 31 = 40, then 40 / 2 = 20, then 20 / 2 = 10. So the two boxes must multiply to 10.

1240 \div 31 \div 2 \div 2 = 10

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 10 into the smallest whole numbers greater than 1. 10 = 2 x 5, and both 2 and 5 are prime, so they cannot be broken down further.

10 = 2 \times 5

10 has only one factor pair other than 1 and 10, namely 2 and 5, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 2 and 5, giving the full prime factorization 1240 = 31 x 2 x 2 x 2 x 5.

1240 = 31 \times 2 \times 2 \times 2 \times 5

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 2 and 5 (so 1240 = 31 x 2 x 2 x 2 x 5).

Review

Multiply back: 31 x 2 x 2 x 2 x 5 = 1240, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 1240 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 10 into the prime factor pair 2 and 5 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 1240 by the given factors to find the leftover 10.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!

Variant 9 answer: The two boxes are 2 and 7 (so 952 = 17 x 2 x 2 x 2 x 7).

The number 952 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$952 = 17 \times 2 \times 2 \times \square \times \square$

Show solution

Understand

Write 952 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 952 = 17 x 2 x 2 x box x box, find the two missing numbers.

Givens

The number is 952
It is already partly factored as 17 x 2 x 2 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 952
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 952 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 952 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 952 by the factors already shown. 952 / 17 = 56, then 56 / 2 = 28, then 28 / 2 = 14. So the two boxes must multiply to 14.

952 \div 17 \div 2 \div 2 = 14

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 14 into the smallest whole numbers greater than 1. 14 = 2 x 7, and both 2 and 7 are prime, so they cannot be broken down further.

14 = 2 \times 7

14 has only one factor pair other than 1 and 14, namely 2 and 7, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 2 and 7, giving the full prime factorization 952 = 17 x 2 x 2 x 2 x 7.

952 = 17 \times 2 \times 2 \times 2 \times 7

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 2 and 7 (so 952 = 17 x 2 x 2 x 2 x 7).

Review

Multiply back: 17 x 2 x 2 x 2 x 7 = 952, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 952 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 14 into the prime factor pair 2 and 7 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 952 by the given factors to find the leftover 14.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!

Variant 10 answer: The two boxes are 3 and 5 (so 2610 = 29 x 2 x 3 x 3 x 5).

The number 2610 is written as a product of the smallest possible whole numbers (other than 1). Fill in each $\square$ with the correct number.

$2610 = 29 \times 2 \times 3 \times \square \times \square$

Show solution

Understand

Write 2610 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 2610 = 29 x 2 x 3 x box x box, find the two missing numbers.

Givens

The number is 2610
It is already partly factored as 29 x 2 x 3 x box x box
Each factor must be a whole number greater than 1, and the factors should be as small as possible (prime)

Unknowns

The two missing factors in the boxes

Constraints

The product of all factors must equal 2610
No factor may be 1

Plan

#7 Identify Subproblems · also uses: #6 Guess and Check

Factoring 2610 fully is one repeated subproblem: divide out the factors that are already given, then keep dividing the leftover by the smallest possible numbers. Dividing 2610 by the known factors leaves a small number whose own factor pair is easy to find.

Execute

#7 Identify Subproblems 4.NBT.B.6

Divide 2610 by the factors already shown. 2610 / 29 = 90, then 90 / 2 = 45, then 45 / 3 = 15. So the two boxes must multiply to 15.

2610 \div 29 \div 2 \div 3 = 15

Each division peels off one known factor, shrinking the problem to a tiny leftover.

#6 Guess and Check 4.OA.B.4

Now split 15 into the smallest whole numbers greater than 1. 15 = 3 x 5, and both 3 and 5 are prime, so they cannot be broken down further.

15 = 3 \times 5

15 has only one factor pair other than 1 and 15, namely 3 and 5, which are already as small as possible.

#7 Identify Subproblems 4.OA.B.4

The two missing factors are 3 and 5, giving the full prime factorization 2610 = 29 x 2 x 3 x 3 x 5.

2610 = 29 \times 2 \times 3 \times 3 \times 5

Collecting all prime pieces gives the smallest-number product.

Answer: The two boxes are 3 and 5 (so 2610 = 29 x 2 x 3 x 3 x 5).

Review

Multiply back: 29 x 2 x 3 x 3 x 5 = 2610, which matches. All factors are prime, so the numbers truly are the smallest possible.

Use a factor tree (Guess and Check, tool 6) starting from 2610 and breaking each composite piece into smaller factors, reaching the same prime factors.

Standards · min grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Breaking 15 into the prime factor pair 3 and 5 and confirming the factors are smallest.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing 2610 by the given factors to find the leftover 15.

💡 This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!