2023 AMC 10A Problems/Problem 23

Revision as of 23:55, 9 November 2023 by Technodoggo (talk | contribs) (Solution 3)

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

Solution 2

We have 4 integers in our problem. Let's call the smallest of them $a$. $a(a+23) =$ either $(a+1)(a+21)$ or $(a+2)(a+22)$. So, we have the following:

$a^2 + 23a = a^2 + 22a +21$ or

$a^2+23a = a^2 + 24a +44$.

The second equation has negative solutions, so we discard it. The first equation has $a = 21$, and so $a + 23 = 44$. If we check $(a+1)(a+21)$ we get $22 \cdot 42 = 21 \cdot 44$. $44$ is $2$ times $22$, and $42$ is $2$ times $21$, so our solution checks out. Multiplying $21$ by $44$, we get $924$ => $9 + 2 + 4 = \boxed{\textbf{(C) 15}}$.

~Arcticturn

Solution 3

From the problems, it follows that

\[x(x+20)&=y(y+23) = N\\
x^2+20x&=y^2+23y\\
4x^2+4\cdot20x &= 4y^2+4\cdot23y\\
4x^2+4\cdot20x+20^2-20^2 &= 4y^2+4\cdot23y+23^3-23^2\\
(2x+20)^2-20^2 &= (2y+23)^2-23^2\\
23^2-20^2 &= (2y+23)^2-(2x+20)^2\\
(23+20)(23-20) &= (2y+23+2x+20)(2y+23-2x-20)\\
43\cdot 3 &= (2y+2x+43)(2y-2x+3)\\
129\cdot 1 &= (2y+2x+43)(2y-2x+3)\\

129 &= (2y+2x+43)\\
86 &= 2y+2x\\
43 &= x+y\\

1 &= 2y-2x+3\\
x-y &= 1\\\] (Error compiling LaTeX. Unknown error_msg)

~Technodoggo

Video Solution 1 by OmegaLearn

https://youtu.be/D_T24PrVk18

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
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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