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    How to find the number of zero in 365! ?
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    Should probably post this in the maths forum...
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    In respect to what?
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    Is this anything to do with binary numbers?
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    87 zeros

    why was I negged?
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    there are 'math error' zeros in 365.
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    I'll answer this anyway...

    Think about how many multiples of fives there are in 365!, remember 25 and multiples and 125 and multiples.
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    the method is
    you need both 5 and 2 to make 10 (or 1 zero)

    so start dividing by 365/5 you get 71 as the divisor ignore the remainder then divide 71/5 you get 14 ignore the remainder then 14/5 then 2/5

    so now you add 71+14+2 = 87

    you can check for 2 as well this number will be larger than 87, so we always choose the smaller of the two values as you can only make zero with equal 5 and equal 2
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    (Original post by joy12)
    the method is
    you need both 5 and 2 to make 10 (or 1 zero)

    so start dividing by 365/5 you get 71 as the divisor ignore the remainder then divide 71/5 you get 14 ignore the remainder then 14/5 then 2/5

    so now you add 71+14+2 = 87

    you can check for 2 as well this number will be larger than 87, so we always choose the smaller of the two values as you can only make zero with equal 5 and equal 2
    This method does give you the right answer, given that you can divide.
    But would you like to give an explanation for the method?
    This seems much like just a 'learnt' method for finding the number of zeros of a factorial.
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    (Original post by seanahnuk)
    This method does give you the right answer.
    Strange, I get 166 zeroes, of which 89 are trailing:

    365!=
    25104128675558732292929443748812 02770516552026987607976687259519 39011061382209374196660180090002 54169376172314360982328660708071 12336997985344536791065387238359 97043555327409376780914914294408 64316046925074510134847025546014 09800590796554104119549610531188 61733734351455171932827608477558 82291690213539123479186274701519 39680850494072260703300124632839 88005504874279998766904169734378 61078185344667966871511049653888 13013683619901052918005612584454 94886486176829158263475641489909 84138067809999604687488146734837 34069935983879112499595758453887 36166615330932535512568450560463 88738129702951381151861413688922 98651000544094394301469924411255 57552791407604927642537402504103 91056421979003289600000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000
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    Surprising the difficulties people have dividing 365 by 5...
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    (Original post by ghostwalker)
    Strange, I get 166 zeroes, of which 89 are trailing:

    365!=
    25104128675558732292929443748812 02770516552026987607976687259519 39011061382209374196660180090002 54169376172314360982328660708071 12336997985344536791065387238359 97043555327409376780914914294408 64316046925074510134847025546014 09800590796554104119549610531188 61733734351455171932827608477558 82291690213539123479186274701519 39680850494072260703300124632839 88005504874279998766904169734378 61078185344667966871511049653888 13013683619901052918005612584454 94886486176829158263475641489909 84138067809999604687488146734837 34069935983879112499595758453887 36166615330932535512568450560463 88738129702951381151861413688922 98651000544094394301469924411255 57552791407604927642537402504103 91056421979003289600000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000
    Ok mr smarty-pants.

    EDIT: They should change the word negger to nerdier solely for the benefit of the maths forum.
    EDIT2: Oh, and I like being negged, it makes me more determined.
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    Each trailing zero corresponds to a multiple of 10, so if you want to calculate the number of trailing zeros in a number N whose prime factorisation is 2^a5^bM then there will be \min\{a,b\} zeros.

    In a factorial, there will always be more 2s than 5s in the prime factorisation (why?) so you only need to worry about 5s. You get one factor of 5 for each multiple of 5 less than or equal to 365, and an additional 5 for each multiple of 25, and an additional factor of 5 for each multiple of 125. You just need to add all these up and this gives you the number of zeros.
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    (Original post by nuodai)
    ..
    I don't think the issue is method, more that the only person who posted a solution made an arithmetic error.
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    (Original post by DFranklin)
    I don't think the issue is method, more that the only person who posted a solution made an arithmetic error.
    I wasn't responding to those, I was responding to the OP (who presumably hasn't really been helped by the preceding posts).
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    (Original post by seanahnuk)
    This method does give you the right answer, given that you can divide.
    But would you like to give an explanation for the method?
    This seems much like just a 'learnt' method for finding the number of zeros of a factorial.
    I explained with the first sentence, I mentioned clearly that you need both 5 and 2 to make a 10.

    If there are 100 2's and 400 5's then you can only make 200 10's.

    and if you have 20 5's and 100 2's then you can make 20 10's

    so when we divide a number successively by 5 we get to know how many 5's there are in that number and same is the case with 2

    Yes, I learnt the method, but I understood it as well. Majority of numbers will have more 2's than 5's so we go with 5's.


    go on neg me more, don't care anymore! _I_
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    (Original post by joy12)
    I explained with the first sentence, I mentioned clearly that you need both 5 and 2 to make a 10.
    I think the "understanding" question is about why repeated division by 5 (and summing the quotients) gives the number of 5's dividing 365!

    You also still seem unaware that 365/5 is not 71.

    [Not that either is a big deal, but if you're going to be upset about the comments, you might as well be upset for the right reasons].
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    (Original post by DFranklin)
    I think the "understanding" question is about why repeated division by 5 (and summing the quotients) gives the number of 5's dividing 365!

    You also still seem unaware that 365/5 is not 71.

    [Not that either is a big deal, but if you're going to be upset about the comments, you might as well be upset for the right reasons].
    Thanks for pointing it out

    My apologies, I made a mistake it should be 73 and not 71.
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    89 zero.... in 365! ...
    Way to find out trailing zero
    365/5= 73
    365/25= 14{approx)
    365/125= 2(approx)
    73+14+2= 89
    So there are 89 zero trailing with 365!
 
 
 
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