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Math Help - modular arith

  1. #1
    Senior Member sfspitfire23's Avatar
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    modular arith

    What is the multiplication of 67th and 68th term in the sequence 3, 4, 2, 1, 3, 4, 2, 1, 3, 4......?

    Is it possible to solve with mod arithmetic? So, sequence starts over every 4 numbers, so take mod 4. Then, 67 is equivalent to 3 (mod 4) and 68 is equivalent to 0 (mod 4). So we have 1(3)=3?

    This seems off to me.
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  2. #2
    Super Member Bacterius's Avatar
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    (nevermind this, didn't properly see the sequence). Yes, this is right in my opinion. It makes sense.
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  3. #3
    Senior Member sfspitfire23's Avatar
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    hmm. i think the real answer is 1(2)=2 though. Another way to do it would be to recognize that the 64th term will be 1 and so the 67th = 2 and 68th=1....
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    hmm. i think the real answer is 1(2)=2 though. Another way to do it would be to recognize that the 64th term will be 1 and so the 67th = 2 and 68th=1....
    Yes it is 2*1 = 2.

    67 is congruent to 3 (mod 4) means, use the 3rd term, which is 2.
    68 is congruent to 0 (mod 4) means, use the 4th term, which is 1.

    There is no 0th term so you use the 4th term.
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