Problem 18 · 2026 AMC 8
Hard
Number Theory
consecutive-sums
In how many ways can 60 be written as the sum of two or more consecutive odd positive integers, arranged in increasing order?
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Answer: B — 2.
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Hint 1 of 2
Before any algebra, use parity: a sum of k odd numbers is even exactly when k is even. Since 60 is even, that already says something about how many terms you can have.
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Hint 2 of 2
A run of k consecutive odd numbers starting at a sums to k(a + k − 1) = 60. Test only the even k that divide 60, keeping those with a positive odd start.
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Approach: parity narrows k, then the run-sum formula finishes it
- First a free filter: adding k odd numbers gives a result with the same parity as k. Our target 60 is even, so the number of terms k must be even — we never even test odd k.
- A run of k consecutive odds starting at a is a + (a+2) + … = k(a + k − 1) = 60, so a = 60/k − (k − 1). Check the even k ≥ 2 dividing 60:
- k = 2 → a = 29 (29 + 31 = 60); k = 6 → a = 5 (5+7+9+11+13+15 = 60); k = 4, 10, 12… give a ≤ 0 or non-odd. Two runs work.
- So there are 2 ways.
- Why this transfers: on ‘sum of consecutive’ problems, a parity (or mod) check on the target usually kills most cases for free before you grind the algebra — cheap filters first, formula second.
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