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How many $0$’s are there at the end of $50!$?
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We get a 0 at end of a number if the number is divisible by 10. In terms of prime numbers this means a number being divisible by each pair of (2, 5) we get a 0 at end of it. When we consider factorial, number of 2's will always be greater than number of 5's. So, we just need to count the no. of powers of 5.

Number of $0$ at the end of $50!$ will be = $\frac{50}{5}$ + $\frac{50}{25}$ + $\frac{50}{125}$

                                                             = 10 + 2 + 0

                                                             = 12
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as we know for divisibilty of 10..any number must be divisible by  2 and 5...so whenever we want no of 0's at the end of any number..it will simply show how much time we can we can divide this number by 10...now The idea is to consider prime factor of a factorial n.If we can count the number of 5s and 2s, our task is done.Consider the following examples.

n = 5: There is one 5 and 3 2s in prime factors of 5! =(2 * 2 * 2 * 3 * 5).no. of zero=1

n = 11: There are two 5s and three 2s in prime factors of 11! (2 8 * 34 * 52 * 7). no of zero at the end=2

We can easily observe that the number of 2s in prime factors is always more than or equal to the number of 5s. So if we count 5s in prime factors, we are done. now 

 Count of 5s in prime factors of n!
                  = floor(n/5) + floor(n/25) + floor(n/125) + ....
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50!/5=10

10/5=2

so 50!end with 12 zeros
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When we consider factorial, number of 2's will always be greater than number of 5's. So, we just need to count the no. of powers of 5.

In n! , we have factors that are multiples of 25,125….. If we divide them by 5(25, 50, 75, 100)/5 we get (5, 10, 15, 20), still multiples of 5. This means that we still have  5's that needs an even pair, so we need to divide again by 5.

Now, since we have already counted the 5’s in the first division, we need to count the second set of 5’s. Instead of dividing again by 5, we divide them by 25 and next by 125 and so on. That is the reason why we also divide n by powers of 5.

 Count of 5s in n! = floor(n/5) + floor(n/25) + floor(n/125) + .…

                                       =   $\frac{50}{5}$ +$\frac{50}{25}$ + $\frac{50}{125}$

                                       =  10  +  2  +  0

                                       = 12

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