2021 AMC 12B Fall 第 16 题

By symmetry assume $a\le b\le c;$ then $a\le7$ and $c=23-a-b$

Use $\gcd(b,c)=\gcd(b,23-a)$ and check the short range $a\le b\le\frac{23-a}{2}$

The condition is symmetric, so assume $a\le b\le c.$ Then $a\le7,$ and for each fixed $a$ we have $a\le b\le\lfloor\frac{23-a}{2}\rfloor$ and $c=23-a-b.$ The gcd sum can be evaluated as $$\begin{aligned} &\gcd(a,b)+\gcd(b,23-a) \\ &\quad {}+\gcd(a,23-b). \end{aligned}$$

Checking these short ranges, the possible gcd sums for $a=1,2,\ldots,7$ are, respectively, $\{3,4,13\},$ $\{4,6,10\},$ $\{3,5,6,8,9,12\},$ $\{4,6\},$ $\{3,7,8,11\},$ $\{4,6,8\},$ and $\{9,10\}.$ The value $9$ occurs only at $(a,b,c)=(3,5,15)$ and $(7,7,9).$

The triple $(7, 7, 9)$ has $\gcd$ sum $7 + 1 + 1 = 9$ and $a^2 + b^2 + c^2 = 49 + 49 + 81$ $= 179.$ The triple $(3, 5, 15)$ has $\gcd$ sum $1 + 5 + 3 = 9$ and $a^2 + b^2 + c^2 = 9 + 25$ $+ 225 = 259.$

The sum of the distinct values is $179 + 259 = 438.$

Thus, the correct answer is B.