Question

Suppose $a,$ $b,$ and $c$ are positive integers such that

$\frac{a}{14}+\frac{b}{15}=\frac{c}{210}.$

Which of the following statements are necessarily true?

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

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

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

I, II, and III

I only

I and II only

III only

II and III only

Solution:

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

Thus, the correct answer is E.

2023 AMC 12B Problem 15

Solution:

Problem 15 in Other Years