What is the highest $n$ such that $15^n\mid 100!\;?$ Can anybody please solve this problem? It's really confusing.
 A: $\underline{\text{Hint}}$: Find the highest $n$ such that $5^n\mid 100!$.
$\underline{\text{Food for thought}}$: $\large\lfloor \frac{100}{5}\rfloor +\lfloor \frac{100}{5^2}\rfloor+\lfloor \frac{100}{5^3}\rfloor +\cdots $
A: The highest power of 5 that devides 100! is: Note that $5^2$ devides 4 of the numbers. And there are 20-4=16 numbers that is divided by $5$ but not by $5^2$, thus the highest power of 5 that divides 100! is 2*4+16=24. Note that the highest power of 3 that divides 100! is higher than 24 (the exact number can be calculated similar to the case of 5). Thus 24 is the highest power of 15 that divides 100!
A: Consider what user61527 said in the comments above.
15 will divide $k!$ if the factorization of $k!$ contains a 5 and a 3 (prime factorization of 15).  Thus 5 is the smallest $k$ such that 15 will divide $k!$, seeing as $5!$ factors as $\mathbf{5}*(2*2)*\mathbf{3}*2$.  
$15^2$ will divide $k!$ if the factorization of $k!$ contains two 5's and two 3's.  You don't get another 5 to work with in factoring $k!$ until $k=10$, since $10$ factors as $5*2$, meaning that the smallest $k$ that $15^2$ divides evenly is 10.  Using similar logic, the largest $n$ for which $15^n\mid 100!$ will be the number of (5*3)'s in the prime factorization of $100!$
Now obviously you're going to have a lot more 3's to work with than 5's, since you get a new one for every three numbers, so if we just count the 5's in the prime factorization of $100!$, we can be sure there will be a matching 3 for each of them.
There will be one 5 in the prime factorization of every multiple of 5, plus an additional 5 in the prime factorization of 25, 50, 75, and 100.  That's 24 fives in all, which suggests that 24 is the largest $n$ such that $15^n\mid100!$
