2024 AMC 10A 第 10 题

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10.

考虑如下操作。给定正整数 nn,若 nn33 的倍数,就把 nn 替换为 n3\tfrac{n}{3}。若 nn 不是 33 的倍数,就把 nn 替换为 n+10n + 10。然后继续这个过程。例如,从 n=4n = 4 开始,得到 4142484 \to 14 \to 24 \to 8 186212\to 18 \to 6 \to 2 \to 12 \to \cdots

如果从 n=100n = 100 开始,恰好执行这个操作 100100 次后得到什么值?

Consider the following operation. Given a positive integer n,n, if nn is a multiple of 3,3, then you replace nn by n3.\tfrac{n}{3}. If nn is not a multiple of 3,3, then you replace nn by n+10.n + 10. Then continue this process. For example, beginning with n=4,n = 4, this procedure gives 4142484 \to 14 \to 24 \to 8 186212.\to 18 \to 6 \to 2 \to 12 \to \cdots.

Suppose you start with n=100.n = 100. What value results if you perform this operation exactly 100100 times?

1010

2020

3030

4040

5050

答案:C
知识点:递推模运算过程模拟找规律
难度评级:1350
解答:

100100 开始直接运行:10011012040100 \to 110 \to 120 \to 40 506020\to 50 \to 60 \to 20 301020\to 30 \to 10 \to 20 3010\to 30 \to 10 \to \cdots。第 88 步后到达 1010,之后按 10,20,3010, 20, 30 循环,周期为 33。因此第 8+k8 + k 步是循环中的第 kk 项。对第 100100 步,有 k=92k = 92,且 922(mod3)92 \equiv 2 \pmod 3,所以落在 3030。因此正确答案是 C

Just run it from 100:100: 10011012040100 \to 110 \to 120 \to 40 506020\to 50 \to 60 \to 20 301020\to 30 \to 10 \to 20 3010.\to 30 \to 10 \to \cdots. After the 88th step we're at 10,10, and from there it cycles 10,20,3010, 20, 30 with period 3.3. So step 8+k8 + k is the kkth entry of the cycle. For step 100,100, k=92,k = 92, and 922(mod3),92 \equiv 2 \pmod 3, which lands on 30.30. Therefore, the answer is C.

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