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Math Help - Congruence problems

  1. #1
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    Congruence problems

    I got an assignment from my math teacher to solve some congruence problems, thing is, I have a really hard time understanding how to solve them . So, help from you guys would be greatly appreciated.

    Here are problems:

    1.

    Prove that 3246(mod 11) = 3246(mod 3) + 3246(mod 5)

    2.

    Calculate and simplify:

    1^3 + 2^3 + ... + 13^2 + 14^2 (mod 7)

    Any = should be read as the congruence symbol, don't know how to type that one .
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  2. #2
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    Hello, Currybullen!

    2. Calculate and simplify: . 1^3 + 2^3 + \hdots + 13^2 + 14^2 \text{ (mod 7)}
    We have:

    . . \begin{array}{ccc}1^3 &\equiv& 1^3 \\<br />
2^3 &\equiv&2^3 \\ 3^3 &\equiv&3^3 \\ 4^3 &\equiv& (\text{-}3)^3 \\ 5^3 &\equiv&(\text{-}2)^3 \\ 6^3 &\equiv& (\text{-}1)^3 \\ 7^3 &\equiv & 0^3 \end{array}\;\;\text{(mod 7)}

    \text{Total} \;\;\:\equiv\quad0\:\;\;\text{(mod 7)}


    We find that the same is true for: . 8^3 + 9^3 + \hdots + 14^3


    Therefore: . 1^3 + 2^3 + 3^3 + \hdots + 14^3 \;\equiv\;0\;\text{ (mod 7)}

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  3. #3
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    Thanks I think I grasp that one now. Anyone know about the first one? I'm not even sure it's formulated in the right way, our teacher can be a bit odd at some times.
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  4. #4
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    Can you please clarify

    Prove that 3246(mod 11) = 3246(mod 3) + 3246(mod 5)
    Well, 3246 \equiv 1 (\bmod 11), 3246 \equiv 0 (\bmod 3), 3246 \equiv 1 (\bmod 5). 1=0+1. QED.

    By simply performing these calculations, it can be easily "proved." I do not see anything deeper in this problem. Any random numbers can be thrown together and possess such properties by pure coincidence. What exactly is it your teacher wants you to prove here? Or do you simply have to prove you know your modular arithmetic?
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  5. #5
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    Well I guess it could be like you're saying since it's only a kind of introductuary course.

    Anyway, big thanks for the help guys, just gotta figure some other congruence problems on my own before the test on wednesday.
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