2023 AMC 10A Problems/Problem 23

Problem

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 $N$ ?

$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution 1

Consider positive $a, b$ with a difference of $20$ . Suppose $b = a-20$ . Then, we have that $(a)(a-20) = c$ . If there is another pair of two integers that multiply to 30 but have a difference of 23, one integer must be greater than $a$ , and one must be smaller than $a-20$ . We can create two cases and set both equal. We have $(a)(a-20) = (a+1)(a-22)$ , and $(a)(a-20) = (a+2)(a-21)$ . 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 $c = 924$ . Adding the digits, we have $9+2+4 = \boxed{\textbf{(C) } 15}$ . -Sepehr2010

Video Solution 1 by OmegaLearn

https://youtu.be/D_T24PrVk18

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2023 AMC 10A Problems/Problem 23

Contents

Problem

Solution 1

Video Solution 1 by OmegaLearn

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