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

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== Solution 5 ==
 
== Solution 5 ==
  
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Since we are given that some pairs of divisors differ by 20 and 23 we can do <math>(x-10)(x+10)=</math>(y-<math>\frac{23}{2})(y+</math>\frac{23}{2})<math>, </math>y=\frac{A}{2}<math> </math>x^2-100=\frac{(a^2-529)}{4}<math>. where </math>x-10 and <math>x-10 are factors are the ones that differ by 20 and y+</math>\frac{23}{2} and y-<math>\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 </math>y=\frac{A}{2}<math> </math>x^2-100=\frac{(A^2-529)}{4}<math>. Multiplying by 4 by both sides and simplifying we get that A^2-4x^2=129</math>. We use difference of squares to get that A+2x=129, A-2x=1. So <math>A=65 and </math>x=32.  
+
Since we are given that some pairs of divisors differ by 20 and 23 we can do <math>(x-10)(x+10)=</math>(y-<math></math>\frac{23}{2})(y+<math>\frac{23}{2})</math>, <math>y=\frac{A}{2}</math> <math>x^2-100=\frac{(a^2-529)}{4}</math>. where <math>(x-10) and </math>(x-10) are factors are the ones that differ by 20 and y+<math>\frac{23}{2} and y-</math>\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 <math>y=\frac{A}{2}</math> <math>x^2-100=\frac{(A^2-529)}{4}</math>. Multiplying by 4 by both sides and simplifying we get that A^2-4x^2=129<math>. We use difference of squares to get that A+2x=129, A-2x=1. So </math>A=65 and <math>x=32.  
  
−
Plugging back x into the original equation we get that c=<math>(42)(22) so c=924</math>. The answer is <math>\boxed{\textbf{(C) 15}}</math>.
+
Plugging back x into the original equation we get that c=</math>(42)(22) so c=924<math>. The answer is </math>\boxed{\textbf{(C) 15}}$.
  
 
== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==

Revision as of 15:04, 10 November 2023

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

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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