AMC12 2024 A

AMC12 2024 A · Q9

AMC12 2024 A · Q9. It mainly tests Primes & prime factorization, Remainders & modular arithmetic.

Let $M$ be the greatest integer such that both $M+1213$ and $M+3773$ are perfect squares. What is the units digit of $M$?

设$M$是最大的整数，使得$M+1213$和$M+3773$都是完全平方数。$M$的个位数是多少？

(A) 1 1

(B) 2 2

(D) 6 6

(E) 8 8

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let $M+1213=P^2$ and $M+3773=Q^2$ for some positive integers $P$ and $Q.$ We subtract the first equation from the second, then apply the difference of squares: \[(Q+P)(Q-P)=2560.\] Note that $Q+P$ and $Q-P$ have the same parity, and $Q+P>Q-P.$ We wish to maximize both $P$ and $Q,$ so we maximize $Q+P$ and minimize $Q-P.$ It follows that \begin{align*} Q+P&=1280, \\ Q-P&=2, \end{align*} from which $(P,Q)=(639,641).$ Finally, we get $M=P^2-1213=Q^2-3773\equiv1-3\equiv8\pmod{10},$ so the units digit of $M$ is $\boxed{\textbf{(E) }8}.$

设$M+1213=P^2$和$M+3773=Q^2$，其中$P,Q$为正整数。从第二个方程减去第一个，然后应用平方差：\[(Q+P)(Q-P)=2560.\]注意到$Q+P$和$Q-P$同奇偶，且$Q+P>Q-P$。我们希望最大化$P$和$Q$，所以最大化$Q+P$并最小化$Q-P$。由此 \begin{align*} Q+P&=1280, \\ Q-P&=2, \end{align*} 从而$(P,Q)=(639,641)$。最后，$M=P^2-1213=Q^2-3773\equiv1-3\equiv8\pmod{10}$，所以$M$的个位数是$\boxed{\textbf{(E) }8}$。

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