A number divides by its factor pieces

4.OA.B.44.NBT.B.6 · take · grade 4

Archetype: Decompose a Number into Parts and Factors · step in a 4-type progression

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 × 2 × 2 × □ × □
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 five 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

Ten 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 = 31 x 40 = 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: 1240 = 8 x 155 = (2 x 2 x 2) x (5 x 31), reaching the same five 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 31, 2, and 2 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!