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

Revision as of 23:20, 9 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 has negative solutions, so we discard it. The first one 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

Video Solution 1 by OmegaLearn

https://youtu.be/D_T24PrVk18

@@ Line 7: / Line 7: @@
 Consider positive <math>a, b</math> with a difference of <math>20</math>. Suppose <math>b = a-20</math>. Then, we have that <math>(a)(a-20) = c</math>. If there is another pair of two integers that multiply to 30 but have a difference of 23, one integer must be greater than <math>a</math>, and one must be smaller than <math>a-20</math>. We can create two cases and set both equal. We have <math>(a)(a-20) = (a+1)(a-22)</math>, and <math>(a)(a-20) = (a+2)(a-21)</math>. Starting with the first case, we have <math>a^2-20a = a^2-21a-22</math>,or <math>0=-a-22</math>, which gives <math>a=-22</math>, which is not possible. The other case is <math>a^2-20a = a^2-19a-42</math>, so <math>a=42</math>. Thus, our product is <math>(42)(22) = (44)(21)</math>, so <math>c = 924</math>. Adding the digits, we have <math>9+2+4 = \boxed{\textbf{(C) } 15}</math>.
 -Sepehr2010
+==Solution 2 ==
+We have 4 integers in our problem. Let's call the smallest of them <math>a</math>. <math>a(a+23) = </math> either <math>(a+1)(a+21)</math> or <math>(a+2)(a+22)</math>. So, we have the following:
+<math>a^2 + 23a = a^2 + 22a +21</math> or
+<math>a^2+23a = a^2 + 24a +44</math>.
+The second has negative solutions, so we discard it. The first one has <math>a = 21</math>, and so <math>a + 23 = 44</math>. If we check <math>(a+1)(a+21)</math> we get <math>22 \cdot 42 = 21 \cdot 44</math>. <math>44</math> is <math>2</math> times <math>22</math>, and <math>42</math> is <math>2</math> times <math>21</math>, so our solution checks out. Multiplying <math>21</math> by <math>44</math>, we get <math>924</math> => $9 + 2 + 4 = \boxed{\textbf{(C) 15}}#.
+~Arcticturn
 == Video Solution 1 by OmegaLearn ==

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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