Medium · Worked puzzle

An even remainder narrows the search

Combine parity and bounds to solve a three-blank equation efficiently.

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

□ × □ − □ × 2 = 62

2 distinct solutions · 3 blanks

Try the board first, or open the hints below at your own pace. No answers are shown until you choose to reveal them.

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 2 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

Move twice the last blank to the other side. What kind of number must the product be?

Read the worked approach

Write the equation as ab = 62 + 2c. The product of two keypad digits is at most 81, while c ranges from zero to nine.
The candidate product is therefore one of 62, 64, 66, 68, 70, 72, 74, 76, 78 or 80. It must be even because twice a whole number is even.
Among those values, only 64 and 72 factor into two digits no larger than nine. They are 8 × 8 and 8 × 9. Factors such as 7 × 10 are outside the keypad range.
A product of 64 leaves 2 to subtract, so c = 1. A product of 72 leaves 10 to subtract, so c = 5. Swapping 8 × 9 to 9 × 8 repeats the same solution.

The final blank is multiplied by two. Entering the whole remainder, rather than half of it, subtracts too much.

Parity is most useful when combined with a size limit. Together they remove most of the trial-and-error.

Reveal all 2 solutions

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

8 × 8 − 1 × 2 = 62

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