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Math Help - last two digits

  1. #1
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    last two digits

    what are the last two digits of  77^{25} * 65^{12} ?
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  2. #2
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    \displaystyle (7*11)^{25}=7^{25}*11^{25}

    \displaystyle 7^5\equiv 07 \ (mod \ 100)

    \displaystyle 11^{11}\equiv 11 \ (mod \ 100)

    \displaystyle (7*11)^{25}=7^{25}*11^{25}=(7)^5*(11^{11})^{2+3}\e  quiv 7^5*11^5\equiv  7*51\equiv 57 \ (mod \ 100)

    Now work on 65
    Last edited by dwsmith; November 29th 2010 at 03:57 PM. Reason: Corrected carelessness
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  3. #3
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    The post above me is not quite correct.

    7^5 \equiv 07 mod 100

    So 7^{25} \equiv (07)^5 \equiv 7 mod 100.

    11^{11} \equiv 11 mod 100

    So 11^{25} \equiv 11^3*11^{22} \equiv 31*11 \equiv 51 mod 100.

    So, 77^{25} \equiv 7*51 \equiv 57 mod 100.

    So the last two digits of 77^{25} are 57, not 31.

    NOW, work on 65.
    Last edited by elemental; November 28th 2010 at 07:25 PM. Reason: formatting
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  4. #4
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    I was thinking about inverses and congruence to 1 for some odd reason.
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  5. #5
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    Quote Originally Posted by earthboy View Post
    what are the last two digits of  77^{25} * 65^{12} ?
    Taking smaller moduli...

    Let a=77^{25}\cdot65^{12}. Note that 100=4\cdot25. We can work separately modulo 25 and then modulo 4.

    Clearly 25 divides 65^{12}, and thus 25 divides 77^{25}\cdot65^{12}. So a=77^{25}\cdot65^{12}\equiv0\equiv25\pmod{25}.

    Both 77 and 65 are congruent to 1 modulo 4. So a=77^{25}\cdot65^{12}\equiv1^{25}\cdot1^{12}=1\equ  iv25\pmod4.

    We combine a\equiv25\pmod{25} and a\equiv25\pmod4 to get a\equiv25\pmod{100}. (Since (25,4)=1.)
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  6. #6
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    Hello, earthboy!

    \text}What are the last two digits of }77^{25}\cdot65^{12} \,?

    77^{25}\cdot65^{25} \:=\:(77\cdot65)^{25}

    . . . . . . . . =\:(5005)^{25}

    . . . . . . . . \equiv\:(05)^{25} \text{ (mod 100)}

    . . . . . . . . \equiv\:25\text{ (mod 100)}

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