AMC 8 2026 Problem 18
AMC8 2026 年第 18 题
奇偶性★★★☆☆
In how many ways can 60 be written as the sum of two or more consecutive odd positive integers that are arranged in increasing order?
有多少种方式可以将 60 表示为两个或更多个连续奇正整数（按递增顺序排列）的和？
- A.1
- B.2
- C.3
- D.4
- E.5
正确答案 · Correct Answer
B
解析 · Solution
解法1
n 个连续奇数（首项为 a）之和 = n(a+n−1) = 60。因为 60 是偶数，n 必须是偶数（奇数个奇数之和为奇数）。枚举 60 的偶因数：
n=2：a = −1 = 29 → 29+31=60 ✓
n=4：a = −3 = 12（偶数，非奇数）✗
n=6：a = −5 = 5 → 5+7+9+11+13+15=60 ✓
n=10：a = −9 = −3（负数）✗
共 2 组解。
设 n 个连续奇数为 a, a+2,..., a+2(n−1)，和为 n×a+n(n−1) = n(a+n−1)。n(a+n−1)=60，n 需整除 60。因 60 为偶数，n 必为偶数。检验偶因数：n=2→a=29 ✓，n=6→a=5 ✓。n=4→a=12(偶)✗，n=10→a=−3(负)✗。共 2 组。
Solution 1
The sum of n consecutive odd integers starting from a is: n(a+n−1) = 60. Since the sum is even, n must be even (because the sum of an odd number of odd integers is odd). Trying even divisors of 60:
n=2: a = − 1 = 29 → 29+31=60 ✓
n=4: a = − 3 = 12 (even, not odd) ✗
n=6: a = − 5 = 5 → 5+7+9+11+13+15=60 ✓
n=10: a = − 9 = −3 (negative) ✗
Answer: 2 possible values of n.
Let the n consecutive odd integers be a, a+2,..., a+2(n−1). Their sum is n×a + n(n−1) = n(a+n−1) = 60. Since n(a+n−1)=60 and both factors are integers, n must divide 60. Testing even divisors (because an odd number of odd terms gives an odd sum, but 60 is even): n=2 (a=29) and n=6 (a=5) both give positive odd starting values. n=4 gives a=12 (even), n=10 gives a=−3 (negative). Answer: 2.
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