AMC10 2023 A

AMC10 2023 A · Q23

AMC10 2023 A · Q23. It mainly tests Divisibility & factors, GCD & LCM.

If the positive integer $c$ has positive integer divisors $a$ and $b$ with $c = ab$, then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $c$. Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ by $23$. What is the sum of the digits of $N$?

如果正整数 $c$ 有正整数除数 $a$ 和 $b$ 满足 $c = ab$，则称 $a$ 和 $b$ 是 $c$ 的\textit{互补除数}。假设正整数 $N$ 有一对互补除数之差为 $20$，另一对互补除数之差为 $23$。求 $N$ 的各位数字之和。

(A) 9 9

(B) 13 13

(D) 17 17

(E) 19 19

Answer

Correct choice: (C)

正确答案：（C）

Solution

Consider positive integers $a, b$ with a difference of $20$. Suppose $b = a-20$. Then, we have $(a)(a-20) = n$. If there is another pair of two integers that multiply to $n$ but have a difference of 23, one integer must be greater than $a$, and the other must be smaller than $a-20$. We can create two cases and set both equal. We have $(a)(a-20) = (a+1)(a-22) \text{ or } (a+2)(a-21).$ Note that if we go further to $(a+3)(a-20)$ and beyond, that would violate the condition that one of the two integers must be smaller than $a-20.$ Starting with the first case, we have $a^2-20a = a^2-21a-22$, or $0=-a-22$, which gives $a=-22$, which is not possible. The other case is $a^2-20a = a^2-19a-42$, so $a=42$. Thus, our product is $(42)(22) = (44)(21)$, so $n = 924$. Adding the digits, we have $9+2+4 = \boxed{\textbf{(C) } 15}$.

考虑差为 $20$ 的正整数对 $a, b$，设 $b = a-20$。则 $n = a(a-20)$。若有另一对乘积为 $n$ 但差为 $23$ 的整数，则一个大于 $a$，另一个小于 $a-20$。可分两种情况：$(a)(a-20) = (a+1)(a-22)$ 或 $(a+2)(a-21)$。注意若进一步到 $(a+3)(a-20)$ 等，会违反一个小于 $a-20$ 的条件。第一种情况：$a^2-20a = a^2-21a-22$，得 $0=-a-22$，$a=-22$，不可能。第二种：$a^2-20a = a^2-19a-42$，得 $a=42$。于是 $n = 42\times22 = 44\times21 = 924$。各位数字和 $9+2+4 = \boxed{\textbf{(C) } 15}$

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