AMC12 2005 A

AMC12 2005 A · Q8

AMC12 2005 A · Q8. It mainly tests Primes & prime factorization, Digit properties (sum of digits, divisibility tests).

Let $A,M$, and $C$ be digits with \[(100A+10M+C)(A+M+C) = 2005.\] What is $A$?

设 $A,M$, 和 $C$ 是数字，使得 \[(100A+10M+C)(A+M+C) = 2005.\] $A$ 是多少？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (D)

正确答案：（D）

Solution

Clearly the two quantities are both integers, so we check the prime factorization of $2005 = 5 \cdot 401$. It is easy to see now that $(A,M,C) = (4,0,1)$ works, so the answer is $\mathrm{(D)}$.

显然这两个量都是整数，因此检查 $2005$ 的素因数分解：$2005 = 5 \cdot 401$。现在很容易看出 $(A,M,C) = (4,0,1)$ 可行，所以答案是 $\mathrm{(D)}$。

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