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Thread: last two digits

  1. #1
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    last two digits

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

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

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

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

    Now work on 65
    Last edited by dwsmith; Nov 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.

    $\displaystyle 7^5 \equiv 07 mod 100$

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

    $\displaystyle 11^{11} \equiv 11 mod 100$

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

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

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

    NOW, work on $\displaystyle 65$.
    Last edited by elemental; Nov 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 $\displaystyle 77^{25} * 65^{12} $ ?
    Taking smaller moduli...

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

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

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

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

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

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

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

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

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

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