Problem 22 · 2022 Math Kangaroo
Stretch
Logic & Word Problems
sum-constraint
Each animal in the picture stands for a whole number greater than zero, and different animals stand for different numbers. The two animals in each column add up to the number written beneath that column. What is the largest possible value of the sum of the four numbers in the top row?
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Answer: C — 20
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Hint 1 of 3
Add the four column sums together: that total equals the top row plus the bottom row.
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Hint 2 of 3
The top row is biggest when the bottom row is as small as possible.
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Hint 3 of 3
All eight animal numbers must be different, so the bottom row cannot just be all 1s - find the smallest different numbers that still fit.
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Approach: make the bottom row as small as the all-different rule allows
- The four column sums add up to 36, and that 36 splits into the top row plus the bottom row, so a smaller bottom row means a bigger top row.
- Every animal stands for a different number, so the four bottom animals must be four different numbers; the smallest four different positive whole numbers are 1, 2, 3, 4 - but each top animal must beat its own bottom partner and also stay different from all the rest.
- Choosing bottom values 1, 3, 5, 7 (top partners 2, 4, 6, 8) keeps all eight numbers different and gives the smallest workable bottom row of 16.
- Then the top row is 36 − 16 = 20, and no allowed arrangement does better, so the answer is C.
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