Tanner wants to fill his swimming pool using two hoses, each of which sprays water at a constant rate. Hose $A$ fills the pool in $a$ hours when used by itself, where $a$ is a positive integer. Hose $B$ fills the pool in $b$ hours when used by itself, where $b$ is a positive integer. When used together, Hose $A$ and Hose $B$ fill the pool in 6 hours. How many different possible values are there for $a$ ?
(A) 5
(B) 6
(C) 9
(D) 10
(E) 12
Together, Hose $A$ and Hose $B$ fill the pool in 6 hours.

Thus, it must take Hose $A$ more than 6 hours to fill the pool when used by itself.

Therefore, $a \geq 7$, since $a$ is a positive integer.

Similarly, it must take Hose $B$ more than 6 hours to fill the pool when used by itself.

Therefore, $b \geq 7$, since $b$ is a positive integer.

When used by itself, the fraction of the pool that Hose $A$ fills in 6 hours is $\frac{6}{a}$.

When used by itself, the fraction of the pool that Hose $B$ fills in 6 hours is $\frac{6}{b}$.

When used together, Hose $A$ and Hose $B$ fill the pool once in 6 hours. Thus, $\frac{6}{a}+\frac{6}{b}=1$.

Since $a \geq 7, b \geq 7$, and both $a$ and $b$ are integers, then we can find values for $a$ and $b$ that satisfy the equation $\frac{6}{a}+\frac{6}{b}=1$ by using systematic trial and error.

For example, let $a=7$. Then $\frac{6}{7}+\frac{6}{b}=1$, or $\frac{6}{b}=1-\frac{6}{7}$, or $\frac{6}{b}=\frac{1}{7}$.

Since $\frac{6}{42}=\frac{1}{7}$, then $b=42$ and $a=7$ is one possible solution to the equation $\frac{6}{a}+\frac{6}{b}=1$.

Compare this to what happens when we let $a=11$.

We have $\frac{6}{11}+\frac{6}{b}=1$, or $\frac{6}{b}=1-\frac{6}{11}$, or $\frac{6}{b}=\frac{5}{11}$, or $5 b=66$.

Since there is no integer value for $b$ that makes $5 b$ equivalent to 66 , then $a=11$ does not give a possible solution.

The possible solutions found by systematic trial and error are shown below.

| $a$ | 7 | 8 | 9 | 10 | 12 | 15 | 18 | 24 | 42 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $b$ | 42 | 24 | 18 | 15 | 12 | 10 | 9 | 8 | 7 |

Any value for $a$ larger than 42 requires $b$ to be smaller than 7 , but we know that $b \geq 7$.

Thus, there are only 9 different possible values for $a$.

Note:

We can reduce the time it takes to complete this trial and error above by recognizing that in the equation $\frac{6}{a}+\frac{6}{b}=1$, the $a$ and $b$ are interchangeable.

That is, interchanging $a$ and $b$ in the equation, gives $\frac{6}{b}+\frac{6}{a}=1$, which is the same equation.

For example, this tells us that since $a=7, b=42$ satisfies the equation, then $a=42, b=7$ satisfies the equation as well.

Moreover, if the pair $(a, b)$ satisfies the equation, then $(b, a)$ satisfies the equation, and if $(a, b)$ does not satisfy the equation, then $(b, a)$ does not satisfy the equation.

This interchangeability of $a$ and $b$ is seen in the symmetry of the list of possible solutions above.

Recognizing that this symmetry must exist allows us to quickly determine the 4 remaining solutions that follow after $a=12, b=12$.

ANswer: (C)