Problem 25 · 2023 AMC 8
Hard
Algebra & Patterns
arithmetic-sequencedivisibilitycasework
Fifteen integers a1, a2, a3, …, a15 are arranged in order on a number line. The integers are equally spaced and have the property that
1 ≤ a1 ≤ 10, 13 ≤ a2 ≤ 20, and 241 ≤ a15 ≤ 250.
What is the sum of the digits of a14?
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Answer: A — 8.
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Hint 1 of 2
‘Equally spaced integers’ means an arithmetic sequence: every step adds the same whole number d. The whole problem turns on pinning down that single d.
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Hint 2 of 2
The span a15 − a1 = 14d. Squeeze it: subtract the smallest a1 from the largest a15 for the upper end and vice-versa, getting 231 ≤ 14d ≤ 249. Only one multiple of 14 lives there.
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Approach: nail d from bounds, then a1, then a14
- Equally spaced = arithmetic, so a fixed integer d is added each step. The clever part: although a1 and a15 are each only known within a window, their difference 14d is squeezed into a narrow range — and that range may contain just one multiple of 14.
- Widest and narrowest gaps: 241 − 10 ≤ 14d ≤ 250 − 1, i.e. 231 ≤ 14d ≤ 249. The only multiple of 14 in there is 238 = 14 × 17, so d = 17.
- Now back-substitute: a2 = a1 + 17 ≤ 20 forces a1 ≤ 3, while a15 = a1 + 238 ≥ 241 forces a1 ≥ 3. The two pincers meet at a1 = 3.
- a14 = a15 − d = (3 + 238) − 17 = 224, so the digit sum is 2 + 2 + 4 = 8. This transfers: when loose bounds multiply into a tight one, a divisibility condition (here ‘multiple of 14’) often leaves a single survivor — squeeze, then sieve.
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