AMC10 2025 A

AMC10 2025 A · Q17

AMC10 2025 A · Q17. It mainly tests GCD & LCM, Remainders & modular arithmetic.

Let $N$ be the unique positive integer such that dividing $273436$ by $N$ leaves a remainder of $16$ and dividing $272760$ by $N$ leaves a remainder of $15$. What is the tens digit of $N$?

设 $N$ 为唯一的正整数，使得 $273436$ 除以 $N$ 余 $16$，$272760$ 除以 $N$ 余 $15$。$N$ 的十位数字是多少？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (E)

正确答案：（E）

Solution

The problem statement implies that $N$ divides both $273436-16=273420$ and $272760-15=272745$. We want to find $N > 16$ that satisfies both of these conditions. Hence, we can just find the greatest common divisor of the two numbers. $\gcd(273420,272745)=\gcd(675,272745)=\gcd(675,45)=45$ by the Euclidean Algorithm, so the answer is $\boxed{\text{(E) }4}.$ Note: If an integer $a$ is congruent to an integer $b$ modulo a positive integer $c$, denoted by $a \equiv b \pmod{c}$, this means that $c$ divides the difference $a-b$

题目表明 $N$ 整除 $273436-16=273420$ 和 $272760-15=272745$。我们要找到满足这两个条件的 $N > 16$。因此，直接求这两个数的最大公因数即可。$\gcd(273420,272745)=\gcd(675,272745)=\gcd(675,45)=45$（用欧几里得算法），因此答案为 $\boxed{\text{(E) }4}$。注：如果整数 $a$ 模正整数 $c$ 同余于整数 $b$，记作 $a \equiv b \pmod{c}$，意为 $c$ 整除差 $a-b$。

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