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 and we can let the pair. We also know the product of both the complementary divisors give the same number so $(x-10)(x+10)=(y-\frac{23}{2})(y+\frac{23}{2})$ . Now we let $y=\frac{a}{2}$ . Then we substitute and get $x^2-100=\frac{(a^2-529)}{4}$ . Finally we multiply by 4 and get $4x^2-a^2=-129, a^2-4x^2=129$ . Then we use differences of squares and get $a$ + $2x$ =129, $a$ - $2x$ =1. We finish by getting $a=$ 65 and $x=32$ . So $(42)(22) = 924$ Adding the digits, we have $9+2+4 = \boxed{\textbf{(C) } 15}$ .

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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