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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 16"

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<math>x=1,y=1,z=7</math>: let <math>a=7A, c=7C,</math> we can try all possibilities of <math>A</math> and <math>C</math> to find that <math>a=7, b=9, c=7</math> is the only solution.  
 
<math>x=1,y=1,z=7</math>: let <math>a=7A, c=7C,</math> we can try all possibilities of <math>A</math> and <math>C</math> to find that <math>a=7, b=9, c=7</math> is the only solution.  
  
−
<math>x=1,y=2,z=6</math>: No solutions. By <math>y</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be even. Therefore, <math>x</math> cannot be equal to <math>1</math>.  
+
<math>x=1,y=2,z=6</math>: No solutions. By <math>y</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>2</math>. Therefore, <math>x</math> cannot be equal to <math>1</math>.  
  
 
<math>x=1,y=3,z=5</math>: C has to be both a multiple of <math>3</math> and <math>5</math>. Therefore, <math>c</math> has to be a multiple of <math>15</math>. The only solution for this is <math>a=5, b=3, c=15</math>.  
 
<math>x=1,y=3,z=5</math>: C has to be both a multiple of <math>3</math> and <math>5</math>. Therefore, <math>c</math> has to be a multiple of <math>15</math>. The only solution for this is <math>a=5, b=3, c=15</math>.  
  
−
<math>x=1,y=4,z=4</math>: No solutions. By <math>y</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be even. Therefore, <math>x</math> cannot be equal to <math>1</math>.  
+
<math>x=1,y=4,z=4</math>: No solutions. By <math>y</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>2</math>. Therefore, <math>x</math> cannot be equal to <math>1</math>.  
  
−
<math>x=2,y=2,z=5</math>: No solutions. By <math>x</math> and <math>y</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be even. Therefore, <math>z</math> cannot be equal to <math>1</math>.  
+
<math>x=2,y=2,z=5</math>: No solutions. By <math>x</math> and <math>y</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>2</math>. Therefore, <math>z</math> cannot be equal to <math>1</math>.  
  
−
<math>x=2,y=3,z=4</math>: No solutions. By <math>x</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be even. Therefore, <math>y</math> cannot be equal to <math>1</math>.  
+
<math>x=2,y=3,z=4</math>: No solutions. By <math>x</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>2</math>. Therefore, <math>y</math> cannot be equal to <math>1</math>.  
  
 
<math>x=3,y=3,z=3</math>: No solutions. As <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>3</math>, <math>a+b+c</math> has to be divisible by <math>3</math>. This contradicts the sum <math>a+b+c=23</math>.  
 
<math>x=3,y=3,z=3</math>: No solutions. As <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>3</math>, <math>a+b+c</math> has to be divisible by <math>3</math>. This contradicts the sum <math>a+b+c=23</math>.  

Revision as of 16:02, 24 November 2021

Problem

Suppose $a$, $b$, $c$ are positive integers such that \[a+b+c=23\] and \[\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9.\] What is the sum of all possible distinct values of $a^2+b^2+c^2$?

$(\textbf{A})\: 259\qquad(\textbf{B}) \: 438\qquad(\textbf{C}) \: 516\qquad(\textbf{D}) \: 625\qquad(\textbf{E}) \: 687$

Solution

Let $\gcd(a,b)=x$, $\gcd(b,c)=y$, $\gcd(c,a)=z$. WLOG, let $x \le y \le z$. We can split this off into cases:

$x=1,y=1,z=7$: let $a=7A, c=7C,$ we can try all possibilities of $A$ and $C$ to find that $a=7, b=9, c=7$ is the only solution.

$x=1,y=2,z=6$: No solutions. By $y$ and $z$, we know that $a$, $b$, and $c$ have to all be divisible by $2$. Therefore, $x$ cannot be equal to $1$.

$x=1,y=3,z=5$: C has to be both a multiple of $3$ and $5$. Therefore, $c$ has to be a multiple of $15$. The only solution for this is $a=5, b=3, c=15$.

$x=1,y=4,z=4$: No solutions. By $y$ and $z$, we know that $a$, $b$, and $c$ have to all be divisible by $2$. Therefore, $x$ cannot be equal to $1$.

$x=2,y=2,z=5$: No solutions. By $x$ and $y$, we know that $a$, $b$, and $c$ have to all be divisible by $2$. Therefore, $z$ cannot be equal to $1$.

$x=2,y=3,z=4$: No solutions. By $x$ and $z$, we know that $a$, $b$, and $c$ have to all be divisible by $2$. Therefore, $y$ cannot be equal to $1$.

$x=3,y=3,z=3$: No solutions. As $a$, $b$, and $c$ have to all be divisible by $3$, $a+b+c$ has to be divisible by $3$. This contradicts the sum $a+b+c=23$.

Putting these solutions together, we have $(7^2+9^2+7^2)+(5^2+3^2+15^2)=179+259=\boxed{\textbf{(B) }438}$

-ConcaveTriangle

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