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

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

$\begin{align*} x(x+20)&=y(y+23) = N\\ x^2+20x&=y^2+23y\\ 4x^2+4\cdot20x &= 4y^2+4\cdot23y\\ 4x^2+4\cdot20x+20^2-20^2 &= 4y^2+4\cdot23y+23^3-23^2\\ (2x+20)^2-20^2 &= (2y+23)^2-23^2\\ 23^2-20^2 &= (2y+23)^2-(2x+20)^2\\ (23+20)(23-20) &= (2y+23+2x+20)(2y+23-2x-20)\\ 43\cdot 3 &= (2y+2x+43)(2y-2x+3)\\ 129\cdot 1 &= (2y+2x+43)(2y-2x+3)\\ \end{align*}$ Since both $(2y+2x+43)$ and $(2y-2x+3)$ must be integer, we get two equations. $\begin{align} 129 or 43 &= (2y+2x+43)\\ 1 or 3&= 2y-2x+3\\ \end{align}$ 43 & 1 yields (0,0) which is not what we want. 129 & 1 yields (22,21) which is more interesting.

Simplifying the equations, we get: $\begin{align*} x+y &= 43\\ x-y &= 1\\ x=22&, y=21\\ N &= (22)(22+20) = 924. \end{align*}$

So, the answer is $\boxed{\textbf{(C) 15}}$ .

~Technodoggo

Solution 4

Say one factorization is $n(n+23).$ The two cases for the other factorization are $(n+1)(n+21)$ and $(n+2)(n+22).$ We know it must be the first because of AM-GM intuition: lesser factors are closer together. Thus, $n(n+23)=(n+1)(n+21)$ and we find that $n=21,c=924$ meaning the answer is $\boxed{\textbf{(C) }15}.$

~DouDragon

Solution 5

Since we are given that some pairs of divisors differ by 20 and 23 we can do $(x-10)(x+10)=$ (y-$$ (Error compiling LaTeX. Unknown error_msg)\frac{23}{2})(y+ $\frac{23}{2})$ , $y=\frac{A}{2}$ $x^2-100=\frac{(a^2-529)}{4}$ . where $(x-10) and$ (x-10) are factors are the ones that differ by 20 and y+ $\frac{23}{2} and y-$ \frac{23}{2} are the ones that differ by 23. Since both divisors are integers y must be in the form of \frac{A}{2} where A is an odd integer. After solving and substituting we get that $y=\frac{A}{2}$ $x^2-100=\frac{(A^2-529)}{4}$ . Multiplying by 4 by both sides and simplifying we get that A^2-4x^2=129 $. We use difference of squares to get that A+2x=129, A-2x=1. So$ A=65 and $x=32.

Plugging back x into the original equation we get that c=$ (Error compiling LaTeX. Unknown error_msg)(42)(22) so c=924 $. The answer is$ \boxed{\textbf{(C) 15}}$.

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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