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  1. #1
    Super Member Showcase_22's Avatar
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    Mods.

    FInd an integer m satisfying $\displaystyle n \leq m \leq n+11$ and $\displaystyle m \equiv 7(mod \ 12)$ in each of the following questions.
    (i) $\displaystyle n=100$
    $\displaystyle 10=(-2) \ mod \ 12$ where $\displaystyle -n<r \leq 0$ for $\displaystyle x \equiv r (mod \ n)$.

    $\displaystyle \Rightarrow 10^2=4 \ mod \ 12$ so I need the number to be $\displaystyle 100+3=103.$

    (ii). $\displaystyle n=10^6$
    $\displaystyle 10=-2 \ mod \ 12$ where $\displaystyle -n<r \leq 0$ for $\displaystyle x \equiv r (mod \ n)$.

    $\displaystyle \Rightarrow \ 10^6=(-2)^6 \ mod \ 12 \Rightarrow 10^6=64 \ mod \ 12 \Rightarrow \ 10^6=4 \ mod \ 12$.

    But I require a number m such that $\displaystyle m=7 \ mod \ 12.$

    Combining my above results reveals the number I need to be $\displaystyle 10^6+3$.

    $\displaystyle (iii). n=-999$
    $\displaystyle -9= 3 \ mod \ 12 \Rightarrow \ -9(111)=3(111) \ mod \ 12 \Rightarrow \ -999=9 \ mod \ 12$

    Comparing this with $\displaystyle 7 \ mod \ 12$ gives $\displaystyle m=-999-2=-1001$.

    My method seems to be working pretty well (i've checked them all on a calculator) but is there an easier way to do them?
    Last edited by Showcase_22; Apr 28th 2009 at 02:44 AM.
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  2. #2
    MHF Contributor

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    Strictly speaking, if the problem is just "find a number, m, such that 0< m< n+ 11 and m= 7 (mod 12)" then "m= 7" is a perfectly valid answer for all n! Did you mean "such that n< m< n+ 11"?

    Any number m, satisfying m= 7 (mod 12) can be written m= 7+ 12k for some integer k. To find such a number such that n< m< n+ 11, divide n by 12 to see how large k must be. For example, with $\displaystyle n= 10^6$, the largest value of n that you have here, 12 divides into [tex]10^6= 1000000[/itex] 83333 times. Taking k= 83333, we have 7+ 12(83333)= 100003.

    For n= -999, divide -999 by 12: 12 divides into -999 -83 times. Taking k= -83, we have 7+ 12(-83)= -989 which satisfies -999< -989< -988.
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  3. #3
    Super Member Showcase_22's Avatar
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    Ah yes I did sorry!!

    One of the problems is that i'm not supposed to use a calculator, I did all the ones in my previous post using mental artihmetic.

    I'm just pretty worried that when I get tested on this I might have a question where instead of $\displaystyle 10^6$ or $\displaystyle 10^2$, I get a number like $\displaystyle 10^{20}$ which is going to be a pain to do!
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