# Zeroes at the end of a FACTORIAL

by murshid_islam
Tags: factorial, zeroes
 P: 361 how can i find the number of zeroes at the end of 100! how can i find the number of zeroes at the end of n! thanks in advance.
P: 57
 Quote by murshid_islam how can i find the number of zeroes at the end of 100! how can i find the number of zeroes at the end of n! thanks in advance.
Your question is the same as finding out how often 5 is a prime factor on the natural numbers.

So I am guessing that 100! has 20 zeros, because in counting by fives, we get to 100 after 20 steps.
HW Helper
P: 1,422
 Quote by SteveRives Your question is the same as finding out how often 5 is a prime factor on the natural numbers. So I am guessing that 100! has 20 zeros, because in counting by fives, we get to 100 after 20 steps.
You are WRONG, SteveRives.
(*)You will have a number which is divisible by 5 for every 5 successive integer.
(*)You will have a number which is divisible by 25 (ie 5.5) for every 25 successive integer.
(*)You will have a number which is divisible by 125 (ie 5.5.5) for every 125 successive integer, and so on...
So you will have 20 numbers which is divisible by 5 (included numbers that is divisible by 25).
And you will have 4 numbers which is divisible by 25.
So you will have: 20 + 4 = 24 zeros in 100!
In general, you will have: n! has:
$$C = \left[ \frac{n}{5} \right] + \left[ \frac{n}{5 ^ 2} \right] + \left[ \frac{n}{5 ^ 3} \right] + ... + \left[ \frac{n}{5 ^ q} \right]$$ zeros at the end.
You can stop the sum at any q such that:
$$\left[ \frac{n}{5 ^ q} \right] = 0$$
Where [...] denotes the integer part of a number. For example : [1.55] = 1, [0.77] = 0, [14.333333] = 14.
Viet Dao,

 P: 57 Zeroes at the end of a FACTORIAL Yes, I just realized that we have to add four more! Why? Because in 20 steps, there are four times that five is reduplicated. Finding the number of times 5 is a prime factor is right, but, for example, when we get to 25 it is in there twice. And so 20 / 5 is 4. There are four times that happens. Add that back into the 20 and we get 24.
 P: 1 The formula given in this thread is inaccurate. It's not the rounding function, but the floor function that should be used. floor(n/5)+floor(n/(52))+floor(n/(53))+... This is because rounding each quotient may occasionally result in counts that are too high. For example: 999! 999/5=199.8~200 (correct value: 199) 999/25=39.96~40 (correct value: 39) 999/125=7.992~8 (correct value: 7) 999/625=1.5984~2 (correct value: 1) Total: 250 (correct value: 246) As you can see, there's an error of 4 zeros! Edit: Never mind. I misinterpreted his symbols. The Rounding function is usually indicated by square brackets. The Floor, or the integer part of the number, is indicated by L-shaped brackets. So naturally without reading the entire post, I jumped to a conclusion.
 P: 1 for writing a program it may be useful to get the number of zeros at the end of a factorial without usage of floor function: n = p(1)*5 + q(1) p(1) = p(2)*5 + q(2) p(2) = p(3)*5 + q(3) etc p(n-1) = p(n)*5 + q(n) so, p(n) = (p(n-1) - q(n))/5 p(n-1) = (p(n-2) - q(n-1))/5 etc where q(i) is actually p(i-1) mod 5 , or q(i) = p(i-1)%5
 P: 1 You provided the easiest solution. One 5 each from 5,10,15,20,30,35,40,45,55.......70,80,85… (excluded 25,50,75,100) so that's 15x1=15 Two 5s each from 25,50,100 Three 5s from 75 15+2x3+3x1=24 It was an easy problem. Why did I think its tough :/ Thanks for answering, everyone.
 Mentor P: 18,290 This thread is 7 years old.

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