2023 AMC 10B Problem 18

Multiply by $210$: $c = 15a + 14b$. Since $210 = 2 \cdot 3 \cdot 5 \cdot 7$, $\gcd(c, 210) = 1$ means none of these four primes divides $c$. Reduce $c$ modulo each:

• mod $2$: $c \equiv 15a \equiv a$; mod $7$: $c \equiv 15a \equiv a$.
• mod $3$: $c \equiv 14b \equiv -b$; mod $5$: $c \equiv 14b \equiv -b$.

So $2$ or $7$ divides $c$ exactly when it divides $a$, and $3$ or $5$ divides $c$ exactly when it divides $b$. Therefore
$$\gcd(c, 210) = 1 \iff \gcd(a, 14) = 1 \text{ and } \gcd(b, 15) = 1,$$
which is statement III. Statement II follows: if $\gcd(c,210) = 1$ then both gcds are $1$, so certainly at least one is.

Statement I fails: take $a = 1$ (so $\gcd(a,14) = 1$) and $b = 3$. Then $c = 15 + 42 = 57 = 3 \cdot 19$ and $\gcd(c, 210) = 3$.

The answer is $\boxed{\textbf{(E)}\ \text{II and III only}}$.

Clearing denominators turns a fraction identity into a linear relation, and a linear relation can be read prime by prime. Because $15 \equiv 1$ mod $2$ and mod $7$ while $14 \equiv 0$, the primes of $14$ "see" only $a$; symmetrically the primes of $15$ see only $b$. Once that split is visible, the three logical statements are settled by the single biconditional, plus one counterexample.

• Accepting statement I by analogy with II without testing it; a = 1, b = 3 gives c = 57, which is divisible by 3.
• Reading II as requiring both gcds to be 1 and rejecting it; "or both" makes II strictly weaker than III, so III true forces II true.