2021 AMC 12B Fall Problem 16

Below is the professionally curated solution for Problem 16 of the 2021 AMC 12B Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Fall solutions, or check the answer key.

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Concepts:greatest common divisorparitycasework

Difficulty rating: 2100

16.

Suppose a,a, b,b, cc are positive integers such that a+b+c=23a + b + c = 23 and gcd(a,b)+gcd(b,c)+gcd(c,a)=9.\gcd(a, b) + \gcd(b, c) + \gcd(c, a) = 9. What is the sum of all possible distinct values of a2+b2+c2?a^2 + b^2 + c^2?

259259

438438

516516

625625

687687

Solution:

The gcd sum of 99 is large, so the numbers share substantial common factors. Searching the partitions of 2323 that meet the condition gives exactly two solution types.

The triple (7,7,9)(7, 7, 9) has gcd\gcd sum 7+1+1=97 + 1 + 1 = 9 and a2+b2+c2=49+49+81=179.a^2 + b^2 + c^2 = 49 + 49 + 81 = 179. The triple (3,5,15)(3, 5, 15) has gcd\gcd sum 1+5+3=91 + 5 + 3 = 9 and a2+b2+c2=9+25+225=259.a^2 + b^2 + c^2 = 9 + 25 + 225 = 259.

The sum of the distinct values is 179+259=438.179 + 259 = 438.

Thus, the correct answer is B.

Problem 16 in Other Years