2007 AMC 10B Problem 25

Combine the fractions:
$$\frac{a}{b} + \frac{14b}{9a} = \frac{9a^2 + 14b^2}{9ab}.$$
This is an integer exactly when $9ab \mid 9a^2 + 14b^2$. Extract consequences using $\gcd(a,b) = 1$.

Bound $b$. From $b \mid 9a^2 + 14b^2$ we get $b \mid 9a^2$, and since $b$ is coprime to $a$, $b \mid 9$. Also $9 \mid 9a^2 + 14b^2$ gives $9 \mid 14b^2$, so $3 \mid b$. Thus $b \in \{3, 9\}$.

Bound $a$. From $a \mid 14b^2$ and coprimality, $a \mid 14$, so $a \in \{1, 2, 7, 14\}$.

Case $b = 9$. The condition becomes $81a \mid 9a^2 + 1134$, so $9 \mid a^2 + 126$, which needs $3 \mid a$. No divisor of $14$ works.

Case $b = 3$. The condition becomes $27a \mid 9a^2 + 126$, i.e. $3a \mid a^2 + 14$. Each $a \in \{1, 2, 7, 14\}$ is coprime to $3$ and divides $14$, and $a^2 + 14 \equiv 1 + 2 \equiv 0 \pmod 3$, so all four work. Directly:
$$\tfrac13 + \tfrac{42}{9} = 5,\quad \tfrac23 + \tfrac{42}{18} = 3,\quad \tfrac73 + \tfrac{42}{63} = 3,\quad \tfrac{14}{3} + \tfrac{42}{126} = 5.$$

The pairs are $(1,3), (2,3), (7,3), (14,3)$: four in all.

The answer is $\boxed{\textbf{(A)}\ 4}$.

"Expression is an integer" becomes a divisibility statement once the fractions are combined, and coprimality converts $b \mid 9a^2$ into $b \mid 9$: a divisor coprime to one factor must divide the other. That squeezes an infinite search into a handful of candidates, which are then checked one by one. The extra factor of $9$ in the denominator is what kills $b = 1$ and $b = 9$.

• Only using b | 9 and a | 14 (twelve candidate pairs) without checking the full divisibility, or overlooking that 9 must divide 14b^2.
• Answering "infinitely many" because scaling $(a, b)$ preserves the expression, forgetting that the coprime condition removes those multiples.