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Math Help - fastest way to solve:

  1. #1
    Senior Member sfspitfire23's Avatar
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    fastest way to solve:

    Given N = 10^{22} + 1. If ‘N’ is divided by 11, then
    Col A: The remainder
    Col B: 2

    This is a gre math Q. Would the fastest way to solve be to recognize that for every even power of 10, there is a remainder of 1 when product is divided by 11? Then the +1 on the end would make remainder 2. Any other ways to tackle this problem?
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Given N = 10^{22} + 1. If ‘N’ is divided by 11, then
    Col A: The remainder
    Col B: 2

    This is a gre math Q. Would the fastest way to solve be to recognize that for every even power of 10, there is a remainder of 1 when product is divided by 11? Then the +1 on the end would make remainder 2. Any other ways to tackle this problem?
    Fastest way is to do 10^{22}+1 \equiv (-1)^{22} + 1 \equiv 1 + 1 \equiv 2 \pmod{11}. Are you familiar with this notation?
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  3. #3
    Senior Member sfspitfire23's Avatar
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    I am familiar with modular arithmetic...however, I don't understand how you got 10^22 to be (-1)^22
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  4. #4
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    10 is one less than 11. Since 11\equiv 0 (mod 11), 10\equiv -1 (mod 11).

    Or, to put it another way, 10+ 1= 11: 10+ 1\equiv 11\equiv 0 (mod 11).

    In modulo 11, 10^{22}\equiv (-1)^{22}\equiv 1 (mod 11).
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  5. #5
    Senior Member sfspitfire23's Avatar
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    Ah right. Now, could I then change the +1 to -10 as 1 is 10 less than 11?

    (-1)^22+-10 is equivalent to 1-10 = -9 is equivalent to 2 (mod 11)? And thus, the remainder is 2? Can we conclude that from method?
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Ah right. Now, could I then change the +1 to -10 as 1 is 10 less than 11?

    (-1)^22+-10 is equivalent to 1-10 = -9 is equivalent to 2 (mod 11)? And thus, the remainder is 2? Can we conclude that from method?
    You could do so, but it adds an extra step.
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