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Math Help - Last three digits

  1. #1
    Senior Member pankaj's Avatar
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    Last three digits

    Find the last three digits of 10!.

    Please do not use congruence modulo method.
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  2. #2
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    Quote Originally Posted by pankaj View Post
    Find the last three digits of 10!.

    Please do not use congruence modulo method.
    It should be obvious that the last two digits with be "00"
     2 \cdot 5 = 10 & x 10 = 100

    That takes care of 1,2,5,10.

    3x4 = 12 using only the last digit with next number
    2x6 = 12
    2x7 = 14
    4x8 = 32
    2x9 = 18

    The last three digits: 800

    10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 3628800

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  3. #3
    Senior Member pankaj's Avatar
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    That is CORRECT.
    Do you have a method in general for n!.
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  4. #4
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    Quote Originally Posted by pankaj View Post
    Do you have a method in general for n!.
    Any factorial greater that 4 will end in zero.
    That should be obvious.
    The last non-zero digit will be 2,4,6, or 8.
    Also obvious.

    You can easily generate a sequence and easily find the last two digits that can occur; then continue that to determine the last three non zero digits.

    It is not difficult and the exercise is good.

    With that information, you can determine the periodic occurence of the last non-zero digit, the last two non-zero digits, the last three non-zero digits, etc...

    Then: you know n, n-1, n-2 , n-3, etc.
    You can map the last sequence of digits to the last digits of the factorial.


    Hope that helps.

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