AMC10 2025 A

AMC10 2025 A · Q12

AMC10 2025 A · Q12. It mainly tests Basic counting (rules of product/sum), Casework.

Carlos uses a $4$-digit passcode to unlock his computer. In his passcode, exactly one digit is even, exactly one (possibly different) digit is prime, and no digit is $0$. How many $4$-digit passcodes satisfy these conditions?

Carlos 使用一个 4 位密码来解锁他的电脑。在他的密码中，正好有一个数字是偶数，正好有一个（可能不同的）数字是质数，且没有数字是 $0$。有多少个 4 位密码满足这些条件？

(A) 176 176

(B) 192 192

(D) 464 464

(E) 608 608

Answer

Correct choice: (D)

正确答案：（D）

Solution

The only two digits that are neither prime nor even are $1$ and $9$. We split this problem into cases based on the number of $2$s. This is because $2$ is both a prime number and an even number. Case 1: For this case, there are no $2$s. There are $4$ choices for where the even digit goes, and $3$ choices for what the even digit is. There are then $3$ choices for where the prime digit goes, and $3$ choices for what the prime number digit is. The last two spots have $2$ choices each, either $1$ or $9$. This gives a total of $4\cdot 3^3 \cdot 2^2 = 432$ options for this case. Case 2: For this case, there is one $2$, with it being the only even prime. There are $4$ choices for where $2$ goes, and $2$ choices for the other three digits each. This case gives a total of $2^3\cdot 4 = 32$ options. Hence, the answer is $432 + 32 = \boxed{\textbf{(D) }464}$

仅有的既非质数又非偶数的两个数字是 $1$ 和 $9$。我们根据 $2$ 的个数分情况讨论。这是因为 $2$ 既是质数又是偶数。情况 1：此情况没有 $2$。偶数字的位置有 $4$ 种选择，偶数字有 $3$ 种选择。然后质数字的位置有 $3$ 种选择，质数字有 $3$ 种选择。最后两个位置各有 $2$ 种选择，要么 $1$ 要么 $9$。此情况总共有 $4\cdot 3^3 \cdot 2^2 = 432$ 种选项。情况 2：此情况有一个 $2$，它是唯一的偶质数。$2$ 的位置有 $4$ 种选择，其余三个数字各有 $2$ 种选择。此情况总共有 $2^3\cdot 4 = 32$ 种选项。因此，答案是 $432 + 32 = \boxed{\textbf{(D) }464}$

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