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Thread: Arithmetic Modular help!!

  1. #1
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    Arithmetic Modular help!!

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

    From my working out, i broke it down to

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


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

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


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

    is this correct?

    thanks!!
    Last edited by jvignacio; Mar 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 $\displaystyle (800)^{367}$ is divided by 73.
    I craniked out the first few powers of 800.

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

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

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


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

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

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

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

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

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

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

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