AMC8 2026

AMC8 2026 · Q16

AMC8 2026 · Q16. It mainly tests Divisibility & factors, Remainders & modular arithmetic.

Consider all positive four-digit integers consisting of only even digits. What fraction of these integers are divisible by $4$?

考虑所有只由偶数组成的正四位数。这些整数中有多少比例是能被 $4$ 整除的？

(A) \frac{1}{4} \frac{1}{4}

(B) \frac{2}{5} \frac{2}{5}

(D) \frac{3}{5} \frac{3}{5}

(E) \frac{3}{4} \frac{3}{4}

Answer

Correct choice: (D)

正确答案：（D）

Solution

The divisibility rule for $4$ is that the last 2 digits must be a multiple of $4$. Because of this, we can "discard" the first two digits, as the last two digits are the only ones that matter (this doesn't actually change the answer, since for every possible case that the first two digits can be, there are the same number of cases the last two digits can be, so the ratio of cases that work to the cases that don't work is still the same). We can also notice that the tens digit also doesn't matter, since every single digit in the number must be even. Therefore, we only need to care about the ones digit. We have the even digits $0$, $2$, $4$, $6$, and $8$, and there are 3 digits that are divisible by $4$, namely $0$, $4$, and $8$, out of $5$ total digits, so we have the final fraction $\boxed{\textbf{(D) }\frac{3}{5}}$.

判断一个数能否被 $4$ 整除的规则是其最后两位数字必须是 $4$ 的倍数。基于此，我们可以“忽略”前两位数字，因为只有最后两位数字决定是否能被 $4$ 整除（这实际上不会改变答案，因为对于前两位数字的每一种可能，最后两位数字的可能情况数量相同，所以满足条件的情况与不满足条件的情况的比例保持不变）。我们还可以注意到，个位数字是关键，因为数字的每位必须是偶数。偶数位分别是 $0$, $2$, $4$, $6$, 和 $8$，其中能被 $4$ 整除的数字有 $0$, $4$ 和 $8$，总共有 $5$ 个偶数。由此，最终分数为 $\boxed{\textbf{(D) }\frac{3}{5}}$。

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