If $n$ is a positive integer, the symbol $n$ ! (read " $n$ factorial") represents the product of the integers from 1 to $n$. For example, $4 !=(1)(2)(3)(4)$ or $4 !=24$. If $x$ and $y$ are integers and $\frac{30 !}{36^{x} 25^{y}}$ is equal to an integer, what is the maximum possible value of $x+y$ ?
(A) 10
(B) 47
(C) 17
(D) 26
(E) 13
For the given expression to be equal to an integer, each prime factor of the denominator must divide out of the product in the numerator.

In other words, each prime number that is a factor of the denominator must occur as a factor at least as many times in the numerator as in the denominator.

We note that $25=5^{2}$ and so $25^{y}=5^{2 y}$.

Also, $36=6^{2}=2^{2} 3^{2}$ and so $36^{x}=\left(2^{2} 3^{2}\right)^{x}=2^{2 x} 3^{2 x}$.

Therefore, the given expression is equal to $\frac{30 !}{2^{2 x} 3^{2 x} 5^{2 y}}$.

We count the number of times that each of 5,3 and 2 is a factor of the numerator.

The product equal to 30 ! includes six factors which are multiples of 5 , namely $5,10,15,20,25,30$.

Each of these factors includes 1 factor of 5 , except for 25 which includes 2 factors.

Therefore, the numerator includes 7 factors of 5 .

For the numerator to include at least as many factors of 5 as the denominator, we must have $7 \geq 2 y$.

Since $y$ is an integer, then $y \leq 3$.

The product equal to 30 ! includes 10 factors which are multiples of 3 , namely $3,6,9,12,15,18,21$, $24,27,30$.

Seven of these include exactly one factor of 3 , namely $3,6,12,15,21,24,30$.

Two of these include exactly two factors of 3 , namely 9,18 .

One of these includes exactly three factors of 3 , namely 27 .

Therefore, the numerator includes $7(1)+2(2)+1(3)=14$ factors of 3 .

For the numerator to include at least as many factors of 3 as the denominator, we must have $14 \geq 2 x$.

Since $x$ is an integer, then $x \leq 7$.

If $x \leq 7$, then the denominator includes at most 14 factors of 2 . Since the product equal to 30 ! includes 15 even numbers, then the numerator includes at least 15 factors of 2 , and so includes more factors of 2 than the denominator. This means that the number of $2 \mathrm{~s}$ does not limit the value of $x$.

Since $x \leq 7$ and $y \leq 3$, then $x+y \leq 7+3=10$.

We note that if $x=7$ and $y=3$, then the given expression is an integer so this maximum can be achieved.

ANSWER: (A)