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Zeroes at the end of a FACTORIAL

  1. Aug 4, 2005 #1
    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.
     
  2. jcsd
  3. Aug 4, 2005 #2
    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.
     
    Last edited: Aug 4, 2005
  4. Aug 4, 2005 #3

    VietDao29

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    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,
     
    Last edited: Aug 4, 2005
  5. Aug 4, 2005 #4
    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.
     
  6. Oct 9, 2010 #5
    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.
     
    Last edited: Oct 9, 2010
  7. Mar 29, 2012 #6
    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
     
  8. Apr 7, 2012 #7
    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.
     
  9. Apr 23, 2012 #8

    micromass

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    This thread is 7 years old.
     
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