# 2021 Fall AMC 12B Problems/Problem 16

## 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 1 (Observation)

Because $a + b + c$ is odd, $a$, $b$, $c$ are either one odd and two evens or three odds.

$\textbf{Case 1}$: $a$, $b$, $c$ have one odd and two evens.

Without loss of generality, we assume $a$ is odd and $b$ and $c$ are even.

Hence, ${\rm gcd} \left( a , b \right)$ and ${\rm gcd} \left( a , c \right)$ are odd, and ${\rm gcd} \left( b , c \right)$ is even. Hence, ${\rm gcd} \left( a , b \right) + {\rm gcd} \left( b , c \right) + {\rm gcd} \left( c , a \right)$ is even. This violates the condition given in the problem.

Therefore, there is no solution in this case.

$\textbf{Case 2}$: $a$, $b$, $c$ are all odd.

In this case, ${\rm gcd} \left( a , b \right)$, ${\rm gcd} \left( a , c \right)$, ${\rm gcd} \left( b , c \right)$ are all odd.

Without loss of generality, we assume $${\rm gcd} \left( a , b \right) \leq {\rm gcd} \left( b , c \right) \leq {\rm gcd} \left( c , a \right) .$$ $\textbf{Case 2.1}$: ${\rm gcd} \left( a , b \right) = 1$, ${\rm gcd} \left( b , c \right) = 1$, ${\rm gcd} \left( c , a \right) = 7$.

The only solution is $(a, b, c) = (7, 9, 7)$.

Hence, $a^2 + b^2 + c^2 = 179$.

$\textbf{Case 2.2}$: ${\rm gcd} \left( a , b \right) = 1$, ${\rm gcd} \left( b , c \right) = 3$, ${\rm gcd} \left( c , a \right) = 5$.

The only solution is $(a, b, c) = (5, 3, 15)$.

Hence, $a^2 + b^2 + c^2 = 259$.

$\textbf{Case 2.3}$: ${\rm gcd} \left( a , b \right) = 3$, ${\rm gcd} \left( b , c \right) = 3$, ${\rm gcd} \left( c , a \right) = 3$.

There is no solution in this case.

Therefore, putting all cases together, the answer is $179 + 259 = \boxed{\textbf{(B)} \: 438}$.

~Steven Chen (www.professorchenedu.com)

## Solution 2 (Enumeration)

Let $\gcd(a,b)=x$, $\gcd(b,c)=y$, $\gcd(c,a)=z$. Without the loss of generality, 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 $1$.

$x=1,y=3,z=5$: Note that $c$ 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 $4$. 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 $5$.

$x=2,y=3,z=4$: 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 $3$.

$x=3,y=3,z=3$: 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

## Solution 3

Since $a+b+c=23$, $\gcd(a,b,c)=23$ or $\gcd(a,b,c)=1$.

As $\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9$, it is impossible for $\gcd(a,b,c)=23$, so $\gcd(a,b,c)=1$.

This means that $\gcd(a,b)$, $\gcd(b,c)$, and $\gcd(c,a)$ must all be coprime. The only possible ways for this to be true are $1+1+7=9$ and $1+3+5=9$.

Without loss of generality, let $a\le b\le c$. Since $a+b+c=23$, then $a=7, b=7, c=9$ or $a=3, b=5, c=15$.

$(7^2+7^2+9^2)+(3^2+5^2+15^2)=179+259=\boxed{\textbf{(B)} \: 438}$.

~bkunzang

~~Hayabusa1