Problem 20 · 2017 Math Kangaroo
Hard
Number Theory
Algebra & Patterns
sum-constraintcasework
Seven positive whole numbers a, b, c, d, e, f, g are written down next to each other in this order. The sum of all seven numbers is 2017. Every two adjacent numbers always differ by 1. Which number can be equal to 286?
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Answer: A — only a or g
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Hint 1 of 2
Neighbours differ by 1, so the seven numbers alternate even, odd, even, … — and the odd total 2017 fixes which positions are even.
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Hint 2 of 2
286 is below the average of about 288, so ask which position can dip the lowest.
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Approach: parity rules out three spots, then check which can reach the smallest value
- Neighbours differ by 1, so the numbers alternate in parity. For the total 2017 to be odd, the four outer-pattern places a, c, e, g must be even and b, d, f odd.
- 286 is even, so it can only sit at one of a, c, e, g — that already rules out choices about b, d, f.
- The average is 2017 ÷ 7 ≈ 288, and since steps are only ±1, the smallest a number can be is 286, reached only by going straight down from an end. An inner even place (c or e) has values on both sides pulling it up, so it cannot get below 288.
- Hence 286 can occur only at an endpoint: only a or g, choice A.
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