2018 AMC 10A Problem 22

Below is the professionally curated solution for Problem 22 of the 2018 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10A solutions, or check the answer key.

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Concepts:greatest common divisorprime factorizationbounding to limit cases

Difficulty rating: 2010

22.

Let a,b,c,a, b, c, and dd be positive integers such that gcd(a,b)=24,\gcd(a, b)=24, gcd(b,c)=36,\gcd(b, c)=36, gcd(c,d)=54,\gcd(c, d)=54, and 70<gcd(d,a)<100.70 < \gcd(d, a) < 100. Which of the following must be a divisor of a?a?

55

77

1111

1313

1717

Solution:

From gcd(a,b)=24=233\gcd(a,b)=24=2^3\cdot3 and gcd(b,c)=36=2232\gcd(b,c)=36=2^2\cdot3^2, the number aa is divisible by 2332^3\cdot3 but not by 323^2.

From gcd(b,c)=36\gcd(b,c)=36 and gcd(c,d)=54=233\gcd(c,d)=54=2\cdot3^3, the number dd is divisible by 2332\cdot3^3 but not by 222^2. Therefore gcd(d,a)=23n\gcd(d,a)=2\cdot3\cdot n, where nn has no factor 22 or 33.

Since 70<6n<10070<6n<100, we have 12<n<1712<n<17. The only possible integer nn with no factor 22 or 33 is 1313, so the factor must be 1313. Thus, D is the correct answer.

Problem 22 in Other Years