Determine the largest positive integer $n$ for which $7^{n}$ is a divisor of the integer $\frac{200 !}{90 ! 30 !}$.

(Note: If $n$ is a positive integer, the symbol $n$ ! (read " $n$ factorial") is used to represent the product of the integers from 1 to $n$. That is, $n !=n(n-1)(n-2) \cdots(3)(2)(1)$. For example, $5 !=5(4)(3)(2)(1)$ or $5 !=120$.)
First, we count the number of factors of 7 included in 200 !.

Every multiple of 7 includes least 1 factor of 7 .

The product 200! includes 28 multiples of 7 (since $28 \times 7=196$ ).

Counting one factor of 7 from each of the multiples of 7 (these are $7,14,21, \ldots, 182,189,196$ ), we see that 200 ! includes at least 28 factors of 7 .

However, each multiple of $7^{2}=49$ includes a second factor of 7 (since $49=7^{2}, 98=7^{2} \times 2$, etc.) which was not counted in the previous 28 factors.

The product 200 ! includes 4 multiples of 49 , since $4 \times 49=196$, and thus there are at least 4 additional factors of 7 in 200 !.

Since $7^{3}>200$, then 200! does not include any multiples of $7^{3}$ and so we have counted all possible factors of 7 .

Thus, 200! includes exactly $28+4=32$ factors of 7 , and so $200!=7^{32} \times t$ for some positive
integer $t$ that is not divisible by 7 .

Counting in a similar way, the product 90 ! includes 12 multiples of 7 and 1 multiple of 49 , and thus includes 13 factors of 7 .

Therefore, $90!=7^{13} \times r$ for some positive integer $r$ that is not divisible by 7 .

Also, 30! includes 4 factors of 7 , and thus $30!=7^{4} \times s$ for some positive integer $s$ that is not divisible by 7 .

Therefore, $\frac{200!}{90!30!}=\frac{7^{32} \times t}{\left(7^{13} \times r\right)\left(7^{4} \times s\right)}=\frac{7^{32} \times t}{\left(7^{17} \times r s\right)}=\frac{7^{15} \times t}{r s}$.

Since we are given that $\frac{200!}{90!30!}$ is equal to a positive integer, then $\frac{7^{15} \times t}{r s}$ is a positive integer. Since $r$ and $s$ contain no factors of 7 and $7^{15} \times t$ is divisible by $r s$, then it must be the case that $t$ is divisible by $r s$.

In other words, we can re-write $\frac{200!}{90!30!}=\frac{7^{15} \times t}{r s}$ as $\frac{200!}{90!30!}=7^{15} \times \frac{t}{r s}$ where $\frac{t}{r s}$ is an integer.

Since each of $r, s$ and $t$ does not include any factors of 7 , then the integer $\frac{t}{r s}$ is not divisible by 7 .

Therefore, the largest power of 7 which divides $\frac{200!}{90!30!}$ is $7^{15}$, and so $n=15$.

ANSWER: 15