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Thread: Congruence

  1. #1
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    Congruence

    Suppose are integers with . For any integer , let . Show that if is congruent to then and .
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  2. #2
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    I will at least rewrite the problem for you. Consider it a favor (since the Ravens are in the playoffs!). Others may now actually look at your problem.


    Suppose $\displaystyle a, b, c, d $ are [non-negative] integers satisfying the conditions
    $\displaystyle 0 \leq a \leq b \leq 99 $
    and
    $\displaystyle 0 \leq c \leq d \leq 99 $.

    For any integer $\displaystyle i \text{, let } n_i = 101i + 1002^i $.


    Show that if
    $\displaystyle n_a + n_b \equiv n_c + n_d \pmod{10100} $
    then
    $\displaystyle a=c \text{ and } b=d $.
    Last edited by abender; Jan 6th 2010 at 06:51 AM.
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  3. #3
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    I'll give you a start.

    First, 10100 is huge. So let's break it down into some of its relatively prime factors. 101, 25, and 4 should do the trick. Note that 101*25*4 = 10100.

    Thus,

    $\displaystyle
    n_a + n_b \equiv n_c + n_d \pmod{10100} \implies
    $

    $\displaystyle
    n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{101}$


    $\displaystyle
    n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{25}$


    $\displaystyle
    n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{4}$
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