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Factorials and unit digit

  1. Apr 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##100!=N\cdot 10^n##. If N is relatively prime with 10 and unit digit of N is d, then n+d is equal to
    A)26
    B)28
    C)30
    D)32


    2. Relevant equations



    3. The attempt at a solution
    I don't think it would be a good idea to expand the factorial and separately write out the factors of 10. I have got no idea about how to approach this problem.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Apr 11, 2013 #2
    I may not be an expert but can you find index power of 10 in 100! ? You must be knowing the method of finding that. Many good course materials like "Vidyamandir" and RSM illustrate that.
     
    Last edited: Apr 11, 2013
  4. Apr 11, 2013 #3

    Ray Vickson

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    In 100!, the multiples of 10 are 10,20,30,...,90, 100, so figuring out what power of 10 these produce is easy. Next, look at all the remaining even factors (2,4,6,8,12,...,18,22...,98) and all the remaining multiples of 5 (5,15,25,35,...,95). Some of these combine to give you factors 2*5 = 10, etc.
     
  5. Apr 11, 2013 #4
    I had thought of this earlier but wouldn't that be a bit tedious? This is a question from my test paper and I suppose there is a easier way. If I go by this method and if I miss even a single factor, I will end up with a wrong answer.
     
  6. Apr 11, 2013 #5

    ehild

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    The numbers from 1-9 produce one zero at the end (2*5). So do the numbers from 11 to 19 (12*15), and so on: 10 zeros.
    The product of 10, 20 ... 100 produce 12 zeros. (100 and 20*50 have to extra zeros)

    Can you increase the number of zeros further at the end of 100! ?

    The remaining numbers are 1,3,46,7,8,9 (mod 10) What is the last number of their product?
    edit: 4 has been left out.

    You have 10 such groups....

    ehild
     
    Last edited: Apr 12, 2013
  7. Apr 11, 2013 #6
    From the 10 groups we have 2 zeroes from the groups: 21-29 and 71-79. This gives us a total of 24 zeroes.

    The last digit of the product of numbers is 2.
    This gives n+d=26.

    Thanks a lot ehild! :smile:

    EDIT: Woops, the unit digit of product of those numbers is 2. There are 10 groups so the unit digit of N is the unit digit of 2^(10) i.e. 4. Hence, the answer is 28.
     
    Last edited: Apr 11, 2013
  8. Apr 12, 2013 #7

    ehild

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    I left out 4 from the remaining numbers.... So the last digit of the product 1*3*4*6*7*8*9 is 8, but it is -2 mod 10, and its 10th power is 4 mod 10, so the result is the same.

    ehild
     
  9. Apr 12, 2013 #8
    Thanks once again ehild! :smile:


    Although I don't understand that modulo arithmetic notation. :tongue2:
     
  10. Apr 13, 2013 #9

    ehild

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    4+10k, k integer:biggrin:
     
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