Problem
A number divides by its factor pieces
Write 1240 as a product of the smallest whole numbers greater than 1 (its prime factors). Given the start 1240 = 31 × 2 × 2 × □ × □, find the two missing numbers.
OperationsBase-ten numbers
Your answer
Watch out It's easy to stop at the leftover 10 without splitting it further — but 10 itself isn't prime, so the boxes need its smallest pieces, 2 and 5.
How to solve
Strategy Identify Subproblems — 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.
1STEP 1
Divide out the known factors
Divide 1240 by 31, then 2, then 2 again: 1240 ÷ 31 ÷ 2 ÷ 2 = 10, so the two boxes must multiply to 10.
1240 ÷ 31 ÷ 2 ÷ 2 = 10
Each division peels off one known factor, shrinking the problem to a tiny leftover.
4.NBT.B.6Identify SubproblemsDividing 1240 by the three factors already shown (31, 2, and 2) leaves exactly the number the two empty boxes must multiply to.
Why?
All five numbers in the row are multiplied together to make 1240, so the three known factors multiplied by the two boxed factors still equals 1240.
Why?
You may multiply the three known factors on their own first and treat their result, 40, as one single factor standing beside the product of the two boxes, because regrouping which factors you combine first never changes the overall product.
🧱Associativity of multiplicationMultiplying three numbers, grouping any two first gives the same product.
Why?
Since 40 multiplied by the two boxes equals 1240, dividing 1240 by 40 hands back exactly what the two boxes multiply to, because division reverses multiplication.
🧱Inverse operationsSubtraction undoes addition and division undoes multiplication — they reverse each other.
2STEP 2
Factor the leftover into smallest pieces
Split 10 into its smallest pieces greater than 1: 10 = 2 × 5, and both are prime.
10 = 2 × 5
Ten has only one factor pair other than 1 and 10, namely 2 and 5, which are already as small as possible.
4.OA.B.4Guess And Check3STEP 3
State the boxes
The missing factors are 2 and 5: the full factorization is 1240 = 31 × 2 × 2 × 2 × 5.
1240 = 31 × 2 × 2 × 2 × 5
Collecting all prime pieces gives the smallest-number product.
4.OA.B.4Identify SubproblemsAnswer
The two boxes are 2 and 5 (so 1240 = 31 × 2 × 2 × 2 × 5).
Multiply back: 31 × 2 × 2 × 2 × 5 = 31 × 40 = 1240, which matches. All factors are prime, so the numbers truly are the smallest possible.
Takeaway
This only needs Grade 4 dividing and factor pairs -- peel off the known factors, then split what is left into the smallest pieces!
- Divide out the known factors
- Factor the leftover into smallest pieces
- State the boxes
Where next?
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Step back and solidifysuggested