zeroes at the end of a FACTORIAL


by murshid_islam
Tags: factorial, zeroes
murshid_islam
murshid_islam is offline
#1
Aug4-05, 02:26 AM
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.
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SteveRives
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#2
Aug4-05, 06:44 AM
P: 57
Quote 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.
VietDao29
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#3
Aug4-05, 07:02 AM
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Quote 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.
24 zeros is the answer.
(*)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:
[tex]C = \left[ \frac{n}{5} \right] + \left[ \frac{n}{5 ^ 2} \right] + \left[ \frac{n}{5 ^ 3} \right] + ... + \left[ \frac{n}{5 ^ q} \right][/tex] zeros at the end.
You can stop the sum at any q such that:
[tex]\left[ \frac{n}{5 ^ q} \right] = 0[/tex]
Where [...] denotes the integer part of a number. For example : [1.55] = 1, [0.77] = 0, [14.333333] = 14.
Viet Dao,

SteveRives
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#4
Aug4-05, 08:07 AM
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.
finkerbean
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#5
Oct9-10, 01:56 PM
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.
devas123
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#6
Mar29-12, 05:19 AM
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
Kishore189
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#7
Apr7-12, 12:55 PM
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.
micromass
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#8
Apr23-12, 11:18 AM
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