Expert · Worked puzzle

Why the divisor must be one

A maximum numerator rules out most of the final blank’s values.

Mathoji.com · 16 October 2026 · Math checked against the game’s rules

□ × (□ + □) ÷ □ = 78

3 distinct solutions · 4 blanks

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Before you play

Each blank takes one digit from 0 to 9. Choices may be repeated.
Follow the printed brackets and the order of operations. Each division must be exact. Division by zero is never allowed.
Find all 3 distinct solutions. Reordering terms of the same addition or multiplication counts once. Subtraction, division and bracket structure still matter.
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Show a starting hint

The numerator is at most 9 × 18. Compare that with multiples of seventy-eight.

Read the worked approach

Rewrite a(b + c) ÷ d = 78 as a(b + c) = 78d. Division by zero is invalid. The numerator can be at most 162, so d can only be one or two.
If d = 2, the numerator would be 156. No factor pair of 156 has one factor at most nine and the other at most eighteen, so this case fails.
For d = 1, the numerator is 78. The only factor pair fitting those bounds is 6 × 13. Therefore a = 6 and b + c = 13.
The three unordered pairs for thirteen are 4 + 9, 5 + 8 and 6 + 7. Put each in the bracket, multiply by six and divide by one to check the target.

Dividing by one is allowed. A seemingly uninteresting digit can be forced by the size of the target.

When the divisor is a blank, bounding the numerator often eliminates most of the search immediately.

Reveal all 3 solutions

One representative of each distinct solution is shown. Other accepted orders of the same addition or multiplication are not repeated.

6 × (9 + 4) ÷ 1 = 78

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