2007 AMC 10B Problem 25

Start with one example to see what can happen. With $a = 1$, $b = 3$:
$$\frac13 + \frac{42}{9} = \frac13 + \frac{14}{3} = \frac{15}{3} = 5,$$
a whole number. So solutions exist. Now find them all.

Step 1: one fraction. Over the common bottom $9ab$,
$$\frac{a}{b} + \frac{14b}{9a} = \frac{9a^2 + 14b^2}{9ab}.$$
This is a whole number exactly when $9ab$ divides $9a^2 + 14b^2$. In particular each of $b$, $a$ and $9$ must divide the top.

Step 2: what can $b$ be?

1. $b$ divides $14b^2$, so for $b$ to divide the top it must divide $9a^2$.
2. $b$ shares no factor with $a$, so it shares none with $a^2$. So $b$ must divide $9$: $b$ is $1$, $3$ or $9$.
3. Also $9$ divides $9a^2$, so $9$ must divide $14b^2$. Since $14$ has no factor $3$, $b^2$ must be a multiple of $9$, so $b$ is a multiple of $3$.
4. That leaves $b = 3$ or $b = 9$.

Step 3: what can $a$ be? In the same way, $a$ divides $9a^2$, so $a$ must divide $14b^2$. It shares no factor with $b$, so $a$ divides $14$: $a$ is $1$, $2$, $7$ or $14$.

Step 4: test the $2 \times 4 = 8$ pairs.

• $b = 9$: the expression is $\frac{a}{9} + \frac{14}{a}$. For $a = 1, 2, 7, 14$ this is $\frac19 + 14$, $\frac29 + 7$, $\frac79 + 2$, $\frac{14}{9} + 1$. The part $\frac{a}{9}$ is never whole and the other part is, so none work.
• $b = 3$: the expression is $\frac{a}{3} + \frac{14}{3a}$.

All four work.

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

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

"This is a whole number" becomes "the bottom divides the top" once the fractions are combined. The key fact is that if a number shares no factor with $a$ and divides $9a^2$, it must divide $9$. That shrinks an endless search to a few candidates you can check by hand.

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 $\boxed{\textbf{(A)}\ 4}$.