Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 3)
(→‎Video Solution by Math-X (First fully understand the problem!!!))
 
(10 intermediate revisions by 9 users not shown)
Line 5: Line 5:
  
 
== Solution 1 ==
 
== Solution 1 ==
−
Consider positive integers <math>a, b</math> with a difference of <math>20</math>. Suppose <math>b = a-20</math>. Then, we have <math>(a)(a-20) = n</math>. If there is another pair of two integers that multiply to <math>n</math> 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>.
+
Consider positive integers <math>a, b</math> with a difference of <math>20</math>. Suppose <math>b = a-20</math>. Then, we have <math>(a)(a-20) = n</math>. If there is another pair of two integers that multiply to <math>n</math> but have a difference of 23, one integer must be greater than <math>a</math>, and the other must be smaller than <math>a-20</math>. We can create two cases and set both equal.  
−
-Sepehr2010
+
 
−
-minor edits by aidanb
+
We have <math>(a)(a-20) = (a+1)(a-22) \text{or} (a+2)(a-21).</math> Note that if we go further to <math>(a+3)(a-20)</math> and beyond, that would violate the condition that one of the two integers must be smaller than <math>a-20.</math>
−
-extremely minor edits by cc21
+
 
−
-no edits in the actual solution by get-rickrolled
+
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, mathboy282, the_eaglercraft_grinder
  
 
==Solution 2 ==
 
==Solution 2 ==
Line 20: Line 22:
 
The second equation has negative solutions, so we discard it. The first equation 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> => <math>9 + 2 + 4 = \boxed{\textbf{(C) 15}}</math>.  
 
The second equation has negative solutions, so we discard it. The first equation 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> => <math>9 + 2 + 4 = \boxed{\textbf{(C) 15}}</math>.  
  
−
~Arcticturn ~Mathkiddie
+
~Arcticturn
  
 
== Solution 3 ==
 
== Solution 3 ==
Line 42: Line 44:
 
1 or 3 &= 2y-2x+3\\
 
1 or 3 &= 2y-2x+3\\
 
\end{align}</cmath>
 
\end{align}</cmath>
−
43 & 1 yields (0,0) which is not what we want.
+
43 & 3 yields (0,0) which is not what we want.
 
129 & 1 yields (22,21) which is more interesting.
 
129 & 1 yields (22,21) which is more interesting.
  
Line 111: Line 113:
  
 
~andliu766
 
~andliu766
 +
 +
==Solution 9==
 +
 +
<math>n(n+20)=m(m+23) \Longrightarrow n^2+20n=m^2+23m \Longrightarrow n^2-m^2+20n-20m=3m</math>. Factoring, <math>(n+m)(n-m)+20(n-m)=3m \Longrightarrow (n-m)(n+m+20)=3m</math>. Let <math>n-m=a>0</math> because clearly <math>n>m</math>. Then. <math>a(2m+20+a)=3m</math>. Note that since <math>20+a>0</math>, if <math>a\geq2</math>, then the equation is <math>4m+a(20+a)>3m</math>, so <math>a-1</math>. Plugging this back, we get <math>2m+21=3m \Longrightarrow m=21</math> and <math>n=22</math>. Now we find <math>N</math> as <math>22*42=924</math> so the answer is <math>15</math>.
 +
 +
-Magnetoninja
 +
 +
==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=qPO3xUAoaBPvvkd2&t=5118
 +
~little-fermat
 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=gxvKNnXX1gjgdkvP&t=8645
 +
 +
~Math-X
 +
 +
==Video Solution ⚡️ 3 min solution ⚡️ ==
 +
 +
https://youtu.be/fuH_b6AieCQ
 +
 +
<i> ~Education, the Study of Everything </i>
  
 
== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==
Line 125: Line 147:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
−
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2023|ab=A|num-b=22|num-a=24}}
 
{{AMC10 box|year=2023|ab=A|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:37, 5 November 2024

Problem

If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$, then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$. 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 integers $a, b$ with a difference of $20$. Suppose $b = a-20$. Then, we have $(a)(a-20) = n$. If there is another pair of two integers that multiply to $n$ but have a difference of 23, one integer must be greater than $a$, and the other must be smaller than $a-20$. We can create two cases and set both equal.

We have $(a)(a-20) = (a+1)(a-22) \text{or} (a+2)(a-21).$ Note that if we go further to $(a+3)(a-20)$ and beyond, that would violate the condition that one of the two integers must be smaller than $a-20.$

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, mathboy282, the_eaglercraft_grinder

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+23^2-23^2\\ (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 & 3 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 of a number are closer together than larger factors of a number. (We can also try both and see which works.) Thus, $n(n+23)=(n+1)(n+21)$ and we find that $n=21,N=924$ 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 be $(x-10)$ and $(x+10)$ as well as $(y-\frac{23}{2})$ and $(y+\frac{23}{2})$ . 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 $9+2+4 = \boxed{\textbf{(C) } 15}$.


~averageguy


Nunber sense note: To avoid tedious multiplication of 2-digit numbers, observe that $n = (42)(22) = (6)(7)(2)(11)$, and $(6)(7)(2) = 84$, and the sum of the digits of $11$ is $2$, so the sum of the digits of $n$ is equivalent to $(8+4)(2) \equiv 24 \equiv 15 \pmod 9$. The only equivalent answer choice is $\boxed{15}$. ~oinava

Solution 6

$N$ can be written $N = \left( a - 10 \right) \left( a + 10 \right)$ with a positive integer $a > 10$ and $N = \left( \frac{2b + 1}{2} - \frac{23}{2} \right) \left( \frac{2b + 1}{2} + \frac{23}{2} \right)$ with a positive integer $b > 11$.

The above equations can be reorganized as \[ \left( 2b + 1 + 2 a \right) \left( 2 b + 1 - 2 a \right) = 43 \cdot 3 . \]

The only solution is $2b + 1 + 2a = 129$ and $2b + 1 - 2a = 1$. Thus, $a = b = 32$. Therefore, $N = 924$. So the sum of the digits of $N$ is $9 + 2 + 4 = \boxed{\textbf{(C)}~15}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 7

We can write $N$ as $a(a+20)$ or $b(b+23)$ where $a$ and $b$ are divisors of $N.$ Since $a(a+20) = b(b+23),$ we know that $a^2 + 20a - b^2 - 23b = 0$, and we can view this as a quadratic in $a.$

Since the solution for $a$ must be an integer, the discriminant for this quadratic must be a perfect square and therefore $20^2 - 4(-b^2 - 23b) = (2c)^2 = 400 + 4b^2 + 92b$ so $b^2 + 23b -c^2 + 100 = 0.$

Since the discriminant of this quadratic in $b$ must also be a perfect square we know that $23^2 - 4(-c^2+100) = d^2$ which we can simplify as $d^2 - 4c^2 = (d-2c)(d+2c) = 129.$ Since they are both positive integers $d - 2c$ and $d + 2c$ are factors of $129 = 3 \cdot 43$ so $d - 2c = 1$ and $d + 2c = 129$ or $d - 2c = 3$ and $d - 2c = 43.$

These systems of equations give us $(c,d) = (32,65)$ and $(c,d) = (10,23)$ respectively, if we plug our values for $c$ into the equation for $b$ we get $b^2 + 23b - 924 = 0$ and $b^2 + 23b = 0$ respectively. The first equation gives us $b = 21$ or $b = -44$ and the second gives us $b = 0$ or $b = -23$, since $b$ is positive we know that $b = 21$ and $N = (21)(21 + 23) = 924$, therefore the sum of the digits of $N$ is $9 + 2 + 4 = \boxed{\textbf{(C) 15}}.$

~SailS

Solution 8 (Trial and Error)

Consider the numbers of the form $a(a+20)$. Since $b(b+23)$ is always even, $a$ is even. Thus, for $a \ge 2$, we calculate $a(a+20)$ for even values of $a$. Then, we check if it can also be represented as a product of numbers that differ by $23$. Checking, we see that $22 \cdot 42 = 21 \cdot 44 = 924$ works. Thus, the answer is $9 + 2 + 4 = \boxed{\textbf{(C) 15}}$

~andliu766

Solution 9

$n(n+20)=m(m+23) \Longrightarrow n^2+20n=m^2+23m \Longrightarrow n^2-m^2+20n-20m=3m$. Factoring, $(n+m)(n-m)+20(n-m)=3m \Longrightarrow (n-m)(n+m+20)=3m$. Let $n-m=a>0$ because clearly $n>m$. Then. $a(2m+20+a)=3m$. Note that since $20+a>0$, if $a\geq2$, then the equation is $4m+a(20+a)>3m$, so $a-1$. Plugging this back, we get $2m+21=3m \Longrightarrow m=21$ and $n=22$. Now we find $N$ as $22*42=924$ so the answer is $15$.

-Magnetoninja

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=qPO3xUAoaBPvvkd2&t=5118 ~little-fermat

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/GP-DYudh5qU?si=gxvKNnXX1gjgdkvP&t=8645

~Math-X

Video Solution ⚡️ 3 min solution ⚡️

https://youtu.be/fuH_b6AieCQ

~Education, the Study of Everything

Video Solution 1 by OmegaLearn

https://youtu.be/D_T24PrVk18

Video Solution by epicbird08

https://youtu.be/HrZ3fia7g2A

~EpicBird08

Video Solution

https://youtu.be/J9VAVT22L40

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_23&oldid=231700"