AMC8 2026

AMC8 2026 · Q18

AMC8 2026 · Q18. It mainly tests Diophantine equations (integer solutions).

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?

有多少种方法可以将 $60$ 写成两个或两个以上递增排列的连续奇正整数的和？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (B)

正确答案：（B）

Solution

Note that the smallest consecutive sum of integers greater than 1 for 10 numbers itself is 55. Thus, we check $k$ consecutive numbers for $2 \le k \le 10$. Case 1: $k=2$ $n+n+1 = 2n+1 = 60$, not possible as $2n+1$ is odd. Case 2: $k=3$ $3n+3 = 60$, in which $n=29$. 1 case. Case 3: $k=4$ $4n+6 = 60$, not possible as 54 is not divisible by 4. Case 4: $k=5$ $5n+10 = 60$ in which $n = 5$, 1 case. Case 6: $k=6$ $6n+15 = 60$, in which 15 is not divisible by 6 and thus no case. Case 7: $k=7$ $7n+21 = 60$, in which 60 isn't divisible by 7 and there are 0 cases. Cases 8-10 continue the same way, with either $k(k+1)/2$ not divisible by $k$ or 60 not divisible by $k$. The answer is $\boxed{2}$.

注意，对于 $k=10$ 个数，连续整数和最小为 55，所以我们检查 $2 \le k \le 10$ 的 $k$ 个连续数的情况。情况1：$k=2$ $n + (n+1) = 2n + 1 = 60$，不可能，因为 $2n+1$ 是奇数。情况2：$k=3$ $3n + 3 = 60$，此时 $n=19$。有 1 种情况。情况3：$k=4$ $4n + 6 = 60$，不可能，因为 54 不能被 4 整除。情况4：$k=5$ $5n + 10 = 60$，此时 $n=10$，有 1 种情况。情况5：$k=6$ $6n + 15 = 60$，15 不能被 6 整除，无情况。情况6：$k=7$ $7n + 21 = 60$，60 不能被 7 整除，无情况。情况7 至情况9 同理，$k(k+1)/2$ 不可被 $k$ 整除或 60 不能被 $k$ 整除。答案是 $\boxed{2}$。

Topics

NT.009 Diophantine equations (integer solutions)