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Math Help - Find last digit

  1. #1
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    Find last digit

    Find the last digit of (a). 13^121, (b). 8^99, (c). 777^777

    I know there's something I do to write these as smaller numbers, but am unsure on how to do this.
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  2. #2
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    The last digit of the result is determined by the exponent and the last digit of the base. Also, for any single digit d, the last digits of d^1, d^2, d^3, ... form a periodic sequence.
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  3. #3
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    Well, the example we did in class was this:
    18^(2011)=(-2)^2011=-2^2011=-2^3*2^2008=-2^3*(2^4)^502=-2^3*6=8*6=8=2(mod10)
    I guess I need to do something like this, but from this example, I don't see where the -2 came from and the mod 10
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  4. #4
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    from this example, I don't see where the -2 came from and the mod 10
    If 0 <= d < 10 and n = d (mod 10), then d is the last digit of n. Also, 18 = -2 (mod 10).
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  5. #5
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    Ok so 13^121
    3^121



    8^99
    (2^3)^99



    777^777
    7^777


    That's as far as I get with all of those
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  6. #6
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    Quote Originally Posted by mathematic View Post
    [SIZE=2]Ok so 13^121
    3^121
    So start making a list.
    (All of these are mod 10)
    3^1 = 3
    3^2 = 9
    3^3 = 7
    3^4 = 1
    3^5 = 3
    etc.

    The sequence of numbers 3, 9, 7, 1 keep repeating. There are four of them, so what is 121 mod 4? That will tell you which of 3, 9, 7, or 1.

    -Dan
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  7. #7
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    13^121
    3^121
    3^1=3 mod 5
    13^121
    1^121 mod 2
    1 mod 2


    8^99
    =0 mod2
    3^99 mod 5
    3^4 mod 5
    =1 mod 5

    777^777
    7^777
    7^2 mod 5
    2^2 mod 5
    =4 mod 4
    =1 mod 2 since 7 is odd
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  8. #8
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    8^99
    =0 mod2
    3^99 mod 5
    3^4 mod 5
    =1 mod 5
    You are correct that 8^{99}\equiv0\pmod{2}. On the other hand, since n^4\equiv1\pmod{5} for all integer 1\le n<5 by Fermat's little theorem, we have 8^{99}\equiv3^{99}\equiv(3^4)^{24}\cdot3^3\equiv3^  3\equiv2\pmod{5}. Now you probably plan to use the Chinese remainder theorem, which says that all solutions to the system of equations x\equiv0\pmod{2}, x\equiv2\pmod{5} are congruent modulo 2\cdot 5. Since 2 and 8^{99} are solutions, 8^{99}\equiv2\pmod{10}.

    One benefit of this approach is that use can use Fermat's little theorem. In contrast, 8^n\not\equiv1\pmod{10} for all n > 0. However, 8^n modulo 10 forms a periodic sequence 8, 4, 2, 6, 8, 4, ... with period 4, as has been said above (and to which you have not responded). Since 99\equiv3\pmod{4}, we conclude that 8^{99} is the 3rd element of the four-element period, i.e., 2. Formally, 8^{99}\equiv(8^4)^{24}\cdot 8^3\equiv6^{24}\cdot2\equiv6\cdot2\equiv2\pmod{10}.
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