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Zeros at the end

  1. Jun 23, 2012 #1
    We want to know how many zeros appear at the end of the number 'N' = 1962!
    In the solution of it,
    We break the number N in a prime numbers notation and which looks like N = 2^a1 x 3^a2 x 5^a3.....p^a(n), so as the question the asks about the zeros in the the product we are concerned about the numbers a1 and a3, as for now because they yield a zero and also about those factors of N which will yield them too.To determine a3, we take that every factor of N divisible by 5 will yield one 5 and so the number of such factors will be 1962/5(and that is my question?) .
    What exactly will the 1962/5 provide us with?
     
  2. jcsd
  3. Jun 23, 2012 #2

    haruspex

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    You also need to consider some of those numbers are divisible by 25, 125...
     
  4. Jun 23, 2012 #3
    Hi kartik
    1962! is 1*2*3*...*1962
    so just look at this sequence of numbers:
    1, 2, 3, 4, 5, ...9, 10, ..., 14, 15, ...1960, 1961, 1962
    As you can see, you have a '5'... every 5 steps :)
    So 1962/5 is how many times you will get a factor 5 and this is a3

    Cheers...
     
  5. Jun 23, 2012 #4

    haruspex

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    No, a3 will be more than that. See my previous post.
     
  6. Jun 23, 2012 #5
    Yes you are correct haruspex, your post wasn't there while I was posting mine
    Indeed, the higher powers of 5 must be considered too, fortunately there are not too many of them and they can be added in by hand
     
  7. Jun 23, 2012 #6
    What haruspex said in general can be formed up as a sum. Say for n!

    [tex]Q = \left [\frac{n}{5} \right ] + \left [ \frac{n}{5^2} \right ] + \left [ \frac{n}{5^3} \right ] + ...... \left [ \frac{n}{5^t} \right ][/tex]

    Q is the number of times the factorial can be divided by 5, and hence 10.

    The sum ends(or each term becomes 0) when 5t exceeds n.

    This in fact, is a particular case to obtain the number of prime divisors of any positive integer factorial.
     
  8. Jun 24, 2012 #7
    " we are concerned about the numbers a1 and a3, as for now because they yield a zero and also about those factors of N which will yield them too"

    I did write about the factors which will also yield a 'n' number of 5s and 2s.

    Infinitum and oli4 are quite convincing, thank you.
     
  9. Jun 24, 2012 #8
    i asked this question once on Yahoo! Answers

    i asked how many trailing zeroes does the factorial of 1 million have?. I did manage to calculate the # of zeroes in Mathematica, but i wanted to know the theory behind it.

    A man answered that you can use "Sterling's asymptotic formula" which is used to approximate factorials at very large values. He managed to find the EXACT number of trailing zeroes i found using Mathematica. i was truly amazed by this :D

    http://en.wikipedia.org/wiki/Asymptotic_formula
     
  10. Jun 24, 2012 #9
    Even for numbers up to 1 million, the sum [itex]\sum\limits_{k=1}^{+\infty} \left\lfloor\dfrac{n}{5^k}\right\rfloor[/itex] only contains 8 terms, namely 200 000, 40 000, 8 000, 1 600, 320, 64, 12, 2, which makes a total of 249 998 trailing zeroes.
     
  11. Jun 24, 2012 #10

    haruspex

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    For powers of 10 it's really easy. You can sum the series to get (10n-2n)/4. I think whoever claimed to use Sterling's formula was kidding.
     
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