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Math Help - Arithmetic Modular help!!

  1. #1
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    Arithmetic Modular help!!

    Hey guys, im trying to find the remainders when this number (800)^{367} is divided into 73.

    From my working out, i broke it down to

    (800)^{367} = (800)^{24*15+7} = (800^{24})^{15} * 800^7


    since 800^7 = (800^4 * 800^3) \equiv 3 remainder (mod 73)
    and since 800^{24} = (800^{12} * 800^{12}) \equiv 1 remainder (mod 73)

    so then im left with (800^{24})^{15} * 800^7 = 1^{15} *3 Reminder (mod 73)


    therefore:
    (800)^{367} \equiv 3 Reminder (mod 73)

    is this correct?

    thanks!!
    Last edited by jvignacio; March 24th 2010 at 06:40 AM.
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  2. #2
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    Any help or ideas guys?
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  3. #3
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    Hello, jvignacio!

    Your answer is correct!

    I approached it differently.
    (No surprise ... There must be brizillions of ways.)


    Find the remainder when (800)^{367} is divided by 73.
    I craniked out the first few powers of 800.

    . . \begin{array}{cccc} 800^1 & \equiv & 70 & \text{(mod 73)} \\<br />
800^2 &\equiv& 9 & \text{(mod 73)} \\<br />
800^3 &\equiv& 46 & \text{(mod 73)} \\<br />
800^4 &\equiv & 8 & \text{(mod 73)} \end{array}

    Then I noted that: . 800^2\cdot800^4 \;\equiv\;9\cdot8 \;\equiv\;72 \text{ (mod 73)}

    Hence: . 800^6 \;\equiv\;-1\text{ (mod 73)}


    We have: . 800^{367} \;=\;800^{6(61)+1}

    . . . . . . . . . . . . =\;800^{6(61)}\cdot800^1

    . . . . . . . . . . . . =\;\left(800^6\right)^{61}\cdot800

    . . . . . . . . . . . . \equiv \;(-1)^{61}\cdot 70 \text{ (mod 73)}

    . . . . . . . . . . . . \equiv\;(-1)(70)\text{ (mod 73)}

    . . . . . . . . . . . . \equiv\; -70\text{ (mod 73)}

    . . . . . . . . . . . . \equiv\;3\text{ (mod 73)}

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