2023 AMC 10B Problem 18

Fractions with different bottoms are awkward, so first clear them.

1. Clear the fractions. Multiply both sides by $210$:
$$15a + 14b = c.$$
2. What "$\gcd(c, 210) = 1$" means. $210 = 2 \cdot 3 \cdot 5 \cdot 7$, so it means none of $2, 3, 5, 7$ divides $c$. In the same way, $\gcd(a, 14) = 1$ means neither $2$ nor $7$ divides $a$, and $\gcd(b, 15) = 1$ means neither $3$ nor $5$ divides $b$.
3. Check $2$ and $7$. Both divide $14b$. So $c = 15a + 14b$ is divisible by $2$ exactly when $15a$ is, and since $15$ has no factor $2$, exactly when $a$ is. The same reasoning works for $7$.
So: $2$ or $7$ divides $c$ exactly when it divides $a$.
4. Check $3$ and $5$. Both divide $15a$. So $c$ is divisible by $3$ exactly when $14b$ is, which is exactly when $b$ is (since $14$ has no factor $3$). The same for $5$.
So: $3$ or $5$ divides $c$ exactly when it divides $b$.
5. Put it together. $c$ avoids all of $2, 3, 5, 7$ exactly when $a$ avoids $2$ and $7$ and $b$ avoids $3$ and $5$. In other words:
$$\gcd(c, 210) = 1 \text{ exactly when both } \gcd(a, 14) = 1 \text{ and } \gcd(b, 15) = 1.$$

Now test the three statements.

• III says exactly this, so III is true.
• II: if $\gcd(c, 210) = 1$, then by step 5 both gcds are $1$, so "one or the other or both" is certainly true. II is true.
• I only asks for one of the two gcds to be $1$, which is not enough. Try $a = 1$ and $b = 3$: then $\gcd(a, 14) = 1$, but $c = 15 + 42 = 57 = 3 \cdot 19$, so $\gcd(c, 210) = 3$. I is false.

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

After clearing the fractions, check each prime of $210$ one at a time. The primes of $14$ cannot see $14b$, so they only care about $a$; the primes of $15$ only care about $b$. Once you know that, one exact rule settles all three statements, and a single small example knocks out the false one. For "which statements are always true" questions, try small numbers to hunt for a counterexample.

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