why trailing zeros in factorial

Hovkial,
trailing zeros in a n! is defined as\(n/5+n/5^2+n/5^3\)...... \(n/5^k\)where \(k <=n\)..

I was wondering the use of this in practical applications. I always understand concepts if i can attribute it to some practical application.
thanks

I still do not understand what you mean by the term "trailing zeroes in n factorial".

Your example is that of a series or a sequence. Series and sequences are very important to model events, e.g., in science, business, mathematics, etc and many applied disciplines.

Other than that, I have no clue about your question. Cheers.

If I understood your question correctly I think you are trying to find the number of 0's at the end of the result of the factorial, hence how many copies of 10 you can take out of the result.

Think of it as counting the number of 10's you can divide out of the factorial, 10 = 5*2 so we need one copy of five and one copy of two for each ten. Then the question is just how many copies of fives there are in the number, since there are plenty of twos to take out. Now typically this means every 5 numbers there will be a factor of 10 however don't forget 25 and 125 for example have more than 1 copy of five. To tackle that, we can count the factors in layers.

For 100! there are 20 copies of fives from 5, 10, 15 ... 95, 100. This is our first layer. The second layer is 25, 50, 75, 100. There are 4 numbers with an additional copy of five, so in total we have 20 + 4 = 24 copies of fives. Finally this results in 24 trailing zero's for 100!. If the question is about 500! instead we would need to add a third layer for each multiple of 125. The formula stated above is not useful for application however the logic and steps taken to reach to that conclusion is within GMAT difficulty. Hopefully that answers what you were asking for!