2023 AMC 12B Problem 15

Attempt Problem 15 of the 2023 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12B solutions, or check the answer key.

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15.

Suppose a,a, b,b, and cc are positive integers such that

a14+b15=c210. \frac{a}{14}+\frac{b}{15}=\frac{c}{210}.

Which of the following statements are necessarily true?

I. If gcd(a,14)=1\gcd(a,14)=1 or gcd(b,15)=1\gcd(b,15)=1 or both, then gcd(c,210)=1.\gcd(c,210)=1.

II. If gcd(c,210)=1,\gcd(c,210)=1, then gcd(a,14)=1\gcd(a,14)=1 or gcd(b,15)=1\gcd(b,15)=1 or both.

III. gcd(c,210)=1\gcd(c,210)=1 if and only if gcd(a,14)=gcd(b,15)=1.\gcd(a,14)=\gcd(b,15)=1.

I, II, and III

I only

I and II only

III only

II and III only

Answer: E
Concepts:greatest common divisormodular arithmeticlogical deduction
Difficulty rating: 1800
Solution:

Multiplying by 210210 gives c=15a+14b.c=15a+14b. Since 151(mod14),15\equiv 1\pmod{14}, we get ca(mod14),c\equiv a\pmod{14}, so gcd(c,14)=1\gcd(c,14)=1 iff gcd(a,14)=1.\gcd(a,14)=1. Since 141(mod15),14\equiv -1\pmod{15}, we get cb(mod15),c\equiv -b\pmod{15}, so gcd(c,15)=1\gcd(c,15)=1 iff gcd(b,15)=1.\gcd(b,15)=1. As 210=1415210=14\cdot 15 with gcd(14,15)=1,\gcd(14,15)=1, statement III follows: gcd(c,210)=1\gcd(c,210)=1 iff both hold. Statement II is the forward implication of III, hence true. Statement I is false: if gcd(a,14)=1\gcd(a,14)=1 but gcd(b,15)1,\gcd(b,15)\ne 1, then gcd(c,15)1,\gcd(c,15)\ne 1, so gcd(c,210)1.\gcd(c,210)\ne 1. Only II and III are true.

Thus, the correct answer is E.

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