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

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 a solution.

x=1, y=2, z=6: No solutions. By y and z, we know that a, b, and c have to all be even. 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 even. 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 even. 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 even. 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 $(7^2+9^2+7^2)+(5^2+3^2+15^2)=179+259=\boxed{\textbf{(B) }438}$

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2021 Fall AMC 12B Problems/Problem 16

Problem

Solution