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"

(→‎Solution 5)
(→‎Video Solution by Math-X (First fully understand the problem!!!))
 
(43 intermediate revisions by 27 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
If the positive integer <math>c</math> has positive integer divisors <math>a</math> and <math>b</math> with <math>c = ab</math>, then <math>a</math> and <math>b</math> are said to be <math>\textit{complementary}</math> divisors of <math>c</math>. Suppose that <math>N</math> is a positive integer that has one complementary pair of divisors that differ by <math>20</math> and another pair of complementary divisors that differ by <math>23</math>. What is the sum of the digits of <math>N</math>?
+
If the positive integer <math>n</math> has positive integer divisors <math>a</math> and <math>b</math> with <math>n = ab</math>, then <math>a</math> and <math>b</math> are said to be <math>\textit{complementary}</math> divisors of <math>n</math>. Suppose that <math>N</math> is a positive integer that has one complementary pair of divisors that differ by <math>20</math> and another pair of complementary divisors that differ by <math>23</math>. What is the sum of the digits of <math>N</math>?
  
 
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math>
 
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math>
  
 
== Solution 1 ==
 
== Solution 1 ==
−
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>.
+
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
+
 
 +
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>
 +
 
 +
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 27: Line 32:
 
x^2+20x&=y^2+23y\\
 
x^2+20x&=y^2+23y\\
 
4x^2+4\cdot20x &= 4y^2+4\cdot23y\\
 
4x^2+4\cdot20x &= 4y^2+4\cdot23y\\
−
4x^2+4\cdot20x+20^2-20^2 &= 4y^2+4\cdot23y+23^3-23^2\\
+
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\\
 
(2x+20)^2-20^2 &= (2y+23)^2-23^2\\
 
23^2-20^2 &= (2y+23)^2-(2x+20)^2\\
 
23^2-20^2 &= (2y+23)^2-(2x+20)^2\\
Line 37: Line 42:
 
<cmath>\begin{align}
 
<cmath>\begin{align}
 
129 or 43 &= (2y+2x+43)\\
 
129 or 43 &= (2y+2x+43)\\
−
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 58: Line 63:
 
== Solution 4 ==
 
== Solution 4 ==
  
−
Say one factorization is <math>n(n+23).</math> The two cases for the other factorization are <math>(n+1)(n+21)</math> and <math>(n+2)(n+22).</math> We know it must be the first because of AM-GM intuition: lesser factors are closer together. Thus, <math>n(n+23)=(n+1)(n+21)</math> and we find that <math>n=21,c=924</math> meaning the answer is <math>\boxed{\textbf{(C) }15}.</math>
+
Say one factorization is <math>n(n+23).</math> The two cases for the other factorization are <math>(n+1)(n+21)</math> and <math>(n+2)(n+22).</math> 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, <math>n(n+23)=(n+1)(n+21)</math> and we find that <math>n=21,N=924</math> meaning the answer is <math>\boxed{\textbf{(C) }15}.</math>
  
 
~DouDragon
 
~DouDragon
Line 64: Line 69:
 
== Solution 5 ==
 
== 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 <math>(x-10)(x+10)=(y-\frac{23}{2})(y+\frac{23}{2})</math> .  
+
Since we are given that some pairs of divisors differ by 20 and 23 and we can let the pair be <math>(x-10)</math> and <math>(x+10)</math> as well as <math>(y-\frac{23}{2})</math> and <math>(y+\frac{23}{2})</math> . We also know the product of both the complementary divisors give the same number so <math>(x-10)(x+10)=(y-\frac{23}{2})(y+\frac{23}{2})</math> .  
 
Now we let <math>y=\frac{a}{2}</math>. Then we substitute and get <math>x^2-100=\frac{(a^2-529)}{4}</math>. Finally we multiply by 4 and get <math>4x^2-a^2=-129, a^2-4x^2=129</math>.  
 
Now we let <math>y=\frac{a}{2}</math>. Then we substitute and get <math>x^2-100=\frac{(a^2-529)}{4}</math>. Finally we multiply by 4 and get <math>4x^2-a^2=-129, a^2-4x^2=129</math>.  
 
Then we use differences of squares and get <math>a</math>+<math>2x</math>=129, <math>a</math>-<math>2x</math>=1. We finish by getting <math>a=</math>65 and <math>x=32</math>. So <math>(42)(22) = 924</math> Adding the digits, we have <math>9+2+4 = \boxed{\textbf{(C) } 15}</math>.
 
Then we use differences of squares and get <math>a</math>+<math>2x</math>=129, <math>a</math>-<math>2x</math>=1. We finish by getting <math>a=</math>65 and <math>x=32</math>. So <math>(42)(22) = 924</math> Adding the digits, we have <math>9+2+4 = \boxed{\textbf{(C) } 15}</math>.
 +
 +
 
~averageguy
 
~averageguy
 +
 +
 +
Nunber sense note: To avoid tedious multiplication of 2-digit numbers, observe that <math>n = (42)(22) = (6)(7)(2)(11)</math>, and <math>(6)(7)(2) = 84</math>, and the sum of the digits of <math>11</math> is <math>2</math>,  so the sum of the digits of <math>n</math> is equivalent to <math>(8+4)(2) \equiv 24 \equiv 15 \pmod 9</math>.  The only equivalent answer choice is  <math>\boxed{15}</math>. ~oinava
 +
 +
==Solution 6==
 +
 +
<math>N</math> can be written <math>N = \left( a - 10 \right) \left( a + 10 \right)</math> with a positive integer <math>a > 10</math> and <math>N = \left( \frac{2b + 1}{2} - \frac{23}{2} \right) \left( \frac{2b + 1}{2} + \frac{23}{2} \right)</math> with a positive integer <math>b > 11</math>.
 +
 +
The above equations can be reorganized as
 +
<cmath>
 +
\[
 +
\left( 2b + 1 + 2 a \right) \left( 2 b + 1 - 2 a \right)
 +
= 43 \cdot 3 .
 +
\]
 +
</cmath>
 +
 +
The only solution is <math>2b + 1 + 2a = 129</math> and <math>2b + 1 - 2a = 1</math>.
 +
Thus, <math>a = b = 32</math>.
 +
Therefore, <math>N = 924</math>.
 +
So the sum of the digits of <math>N</math> is <math>9 + 2 + 4 = \boxed{\textbf{(C)}~15}</math>.
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Solution 7==
 +
We can write <math>N</math> as <math>a(a+20)</math> or <math>b(b+23)</math> where <math>a</math> and <math>b</math> are divisors of <math>N.</math> Since <math>a(a+20) = b(b+23),</math> we know that <math>a^2 + 20a - b^2 - 23b = 0</math>, and we can view this as a quadratic in <math>a.</math>
 +
 +
Since the solution for <math>a</math> must be an integer, the discriminant for this quadratic must be a perfect square and therefore <math>20^2 - 4(-b^2 - 23b) = (2c)^2 = 400 + 4b^2 + 92b</math> so <math>b^2 + 23b -c^2 + 100 = 0.</math>
 +
 +
Since the discriminant of this quadratic in <math>b</math> must also be a perfect square we know that <math>23^2 - 4(-c^2+100) = d^2</math> which we can simplify as <math>d^2 - 4c^2 = (d-2c)(d+2c) = 129.</math> Since they are both positive integers <math>d - 2c</math> and <math>d + 2c</math> are factors of <math>129 = 3 \cdot 43</math> so <math>d - 2c = 1</math> and <math>d + 2c = 129</math> or <math>d - 2c = 3</math> and <math>d - 2c = 43.</math>
 +
 +
These systems of equations give us <math>(c,d) = (32,65)</math> and <math>(c,d) = (10,23)</math> respectively, if we plug our values for <math>c</math> into the equation for <math>b</math> we get <math>b^2 + 23b - 924 = 0</math> and <math>b^2 + 23b = 0</math> respectively. The first equation gives us <math>b = 21</math> or <math>b = -44</math> and the second gives us <math>b = 0</math> or <math>b = -23</math>, since <math>b</math> is positive we know that <math>b = 21</math> and <math>N = (21)(21 + 23) = 924</math>, therefore the sum of the digits of <math>N</math> is <math>9 + 2 + 4 = \boxed{\textbf{(C) 15}}.</math>
 +
 +
~SailS
 +
 +
==Solution 8 (Trial and Error)==
 +
Consider the numbers of the form <math>a(a+20)</math>. Since <math>b(b+23)</math> is always even, <math>a</math> is even. Thus, for <math>a \ge 2</math>, we calculate <math>a(a+20)</math> for even values of <math>a</math>. Then, we check if it can also be represented as a product of numbers that differ by <math>23</math>. Checking, we see that <math>22 \cdot 42 = 21 \cdot 44 = 924</math> works. Thus, the answer is <math>9 + 2 + 4 = \boxed{\textbf{(C) 15}}</math>
 +
 +
~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 ==
 
https://youtu.be/D_T24PrVk18
 
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==
 
==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"