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Math Help - exponetial remainder problem

  1. #1
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    Question exponetial remainder problem

    What is the remainder when 7^348+25^605 is divided by 8?

    The regular calculater errors-out
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ceasar_19134 View Post
    What is the remainder when 7^348+25^605 is divided by 8?

    The regular calculater errors-out
    7 = -1 (mod 8) so 7^348 = (-1)^348 (mod 8) = 1 (mod 8)

    25 = 1 (mod 8) so 25^605 = (1)^605 (mod 8) = 1 (mod 8)

    Thus
    7^348+25^605 = 1 + 1 (mod 8) = 2 (mod 8)

    Thus you will have a remainder of 2.

    -Dan
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  3. #3
    dan
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    Quote Originally Posted by ceasar_19134 View Post
    What is the remainder when 7^348+25^605 is divided by 8?

    The regular calculater errors-out
    do you mean [(7^348) + (25^605)]/8 ?


    dan
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  4. #4
    dan
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    using win XP system calc i got ((7^348) + (25^605)) mod 8 = 2

    but i cant prove that to your teacher ^_^

    dan
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    dan
    yeah
    how do you solve it
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  6. #6
    dan
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    Quote Originally Posted by ceasar_19134 View Post
    dan
    yeah
    how do you solve it
    use topsquarks method... or a win XP calc.....^_^


    see yea,

    dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    7 = -1 (mod 8) so 7^348 = (-1)^348 (mod 8) = 1 (mod 8)

    25 = 1 (mod 8) so 25^605 = (1)^605 (mod 8) = 1 (mod 8)

    Thus
    7^348+25^605 = 1 + 1 (mod 8) = 2 (mod 8)

    Thus you will have a remainder of 2.

    -Dan
    Im still not sure on how 7=-1 and 25=1, could someone explain?
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  8. #8
    dan
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    Quote Originally Posted by ceasar_19134 View Post
    Im still not sure on how 7=-1 and 25=1, could someone explain?

    7 does not = -1 what topsquark said was
    7=-1(mod8)
    same with the other
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ceasar_19134 View Post
    Im still not sure on how 7=-1 and 25=1, could someone explain?
    My apologies. I had figured if you had been given this problem that you would have been taught this. Honestly I really don't know any other (good) way to do your problem.

    Let me start with 25 = 1 (mod 8) first, it's easier.

    This is shorthand for saying that the remainder when we divide 25 by 8 is 1. Another example would be 5 = 2 (mod 3). When you divide 5 by 3 you get a remainder of 2.

    Now, when you divide 7 by 8 you get a remainder of, well, 7. So 7 = 7 (mod 8). Where do I get the -1 from? Well, we're playing around with something called an "equivalence relation." In actuality I shouldn't be using the "=" symbol, it should have 3 bars across, not 2. But since LaTeX is still down I'm using it for shorthand.

    Anyway, this -1 business. We can show that a remainder of 7 (mod 8) is equivalent to a remainder -1 (mod 8). Really it isn't that hard. Let's look at 25 = 1 (mod 8) again. When we do 25/8 we pick the largest integer "k" such that 8k < 25, then we get the remainder by r = 25 - 8k. In this case k = 3. BUT what if we took k = 4 and found r = 25 - 8k? Now r = -7. So 25 = 1 (mod 8) = -7 (mod 8).

    This is all a way of leading you into the more general definition of this "modular" number system. When we say 25 = r (mod 8) what we are saying is that r = 25 - 8k, where k is some integer. This defines a set that r can be equal to. Any member of this set is said to be "equivalent a modulo 8."

    Typically we choose positive numbers for r, but to make things easier we occasionally also pick negative numbers, as I did to solve the problem in this thread.

    This really isn't the best introduction to the subject. (I'm not a bad writer, but neither am I a good one! ) But it isn't really that hard, it just takes a little getting used to. You might find this link to be helpful. The first section is probably all that you need, though if you know the algebra the rest of it is decently informative.

    Hope it helps!

    -Dan
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  10. #10
    Senior Member TriKri's Avatar
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    Is it 25 = 1 (mod 8) and not 25 \equiv 1\ (mod\ 8)? Or what's the difference?
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    Grand Panjandrum
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    Quote Originally Posted by TriKri View Post
    Is it 25 = 1 (mod 8) and not 25 \equiv 1\ (mod\ 8)? Or what's the difference?
    There is no difference.

    RonL
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  12. #12
    Senior Member TriKri's Avatar
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    I'm a noob... -_-;;

    But is there some other difference? (I don't want to start a new thread for this though)
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by TriKri View Post
    I'm a noob... -_-;;

    But is there some other difference? (I don't want to start a new thread for this though)
    Only that you can type "=" in plain ASCII, but a special symbol for
    conguency is redundant as its the (mod 8) qualified identifies what the
    symbol is doing. Also the \equiv symbol has other uses as
    well.

    RonL
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  14. #14
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    Hello, ceasar_19134!

    Exactly what course does this question come from?
    If you aren't familiar with modular arithmetic, it's not a fair question.


    What is the remainder when 7^{348}+25^{605} is divided by 8?

    For reference, let N \,=\,7^{348} + 24^{605}

    Any multiple of 1000 is divisible by 8.
    Hence, we are concerned with only the last three digits of N.

    We will indicate " 7^6 ends in 649" with: 7^6 \rightarrow 649.


    [1] We find that 25^n \rightarrow 625 for n \geq 2.


    [2] We find that:
    . . . . . . \begin{array}{ccccc}7^4 \rightarrow 401 \\ 7^8 \rightarrow 801 \\ 7^{12} \rightarrow 201 \\ 7^{16} \rightarrow 601 \\ 7^{20} \rightarrow 001\end{array}

    Hence: . 7^{348} \:=\:\left(7^{340}\right)\left(7^8\right) \;=\;\left(7^{20}\right)^{17}\left(7^8\right) \rightarrow (001)^{17}(801) \rightarrow 801


    Then: . N \;\rightarrow \;625 + 801 \rightarrow 426

    And: . 426 \div 8 \:=\:53,\;\boxed{\text{remainder }2}

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