How many zeros are there in the number $50!$? How many zeros are there in the number $50!$?
My attempt:
The zeros in every number come from the 10s that make up the number. The 10s are, in turn, made up of 2s and 5s.
So: $\frac{50}{5*2} = 5$ zeros?
 A: The number of 0's is equal to the powers of 5 in the expansion of 50!. This is because the prime decomposition of 50! will have more factors of 2 than factors of 5, and whenever we have a factor of 2 and 5 we can combine them and tack on a 0 at the end of the number.
The number of powers of 5 is $\lfloor{\frac{50}{5}}\rfloor + \lfloor\frac{50}{25}\rfloor = 12$, from an application of Legendre's Formula which can be found here
A: number of zeros  in $n!$ equal to number of $5$ in $n!$ =$$\sum_{k=1}^{\infty}\left \lfloor \frac{n}{5^k} \right \rfloor\\=\left \lfloor \frac{50}{5} \right \rfloor+\left \lfloor \frac{50}{5^2} \right \rfloor+\left \lfloor \frac{50}{5^3} \right \rfloor+...=10+2+0=12$$it means 12 zero in front of 50!
A: In answer to your specific question there are not only 12 zeros at the end of the number but an additional 7 zeros occurring amongst the other non-zero digits
A: Much more than $5$, surely. Since $50!$ is divisible by $2\cdot 5 \cdot 10\cdot 20\cdot 30\cdot 40\cdot 50$, and this number already has $6$ zeroes, you can be sure that $50!$ has at least $6$ zeroes.
In fact, you need to count up how many twos and how many fives appears in the factorization of $50!$. Then, the number of zeroes is the smaller of these two numbers.
