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Difference between revisions of "2023 AMC 10A Problems/Problem 23"

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== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==
 
https://youtu.be/D_T24PrVk18
 
https://youtu.be/D_T24PrVk18
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==See Also==
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{{AMC10 box|year=2023|ab=A|num-b=22|num-a=24}}
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{{MAA Notice}}

Revision as of 20:41, 9 November 2023

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 $9+2+4 = 15$ or $\boxed{C}$. -Sepehr2010

Video Solution 1 by OmegaLearn

https://youtu.be/D_T24PrVk18

See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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