Problem 18 · 2025 Math Kangaroo
Hard
Logic & Word Problems
work-backwardsum-constraint
Maria writes the numbers 1, 2, 3, 4, 5, 6 and 7, each exactly once, into the number wall. Each upper box equals the sum of the two boxes just below it. The bottom-left box already holds 6. Which number must she write in the box with the star?
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Answer: D — 4
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Hint 1 of 2
The box sitting on top of the 6 is 6 plus its neighbour, and it can be at most 7.
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Hint 2 of 2
That forces the neighbour, then keep building upward with 1 to 7 each used once.
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Approach: use that no box can exceed 7 to fix the numbers, then build upward
- Every box is the sum of the two below it, and no number is bigger than 7.
- The box above the 6 is 6 + (its right neighbour), so that neighbour must be 1, giving 6 + 1 = 7.
- The leftover numbers 2, 3, 5 must fill the rest; placing the bottom row as 6, 1, 3, 2 gives the next row 7, 4, 5 - and that uses 1 to 7 exactly once.
- The starred middle-top box is 1 + 3 = 4, option D.
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