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Math Help - find remainder

  1. #1
    Senior Member nikhil's Avatar
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    Lightbulb find remainder

    q 1)what is the remainder when 2^2006 is divided by 7
    ans 1) 2^3 con 1 mod 7
    or 2^2004 con 1 mod 7.......................1
    also 2^2 con -3 mod 7.........................2
    by 1 and 2
    or 2^2006 con -3 mod 7
    hence remainder is 4.
    note: con stands for congruence

    can anyone use the same procedure to find the remainder when 32^[(32)^32] is divided by 7.
    my attempt
    32 con 4 mod 7
    or 32^32 con 4^32 mod 7
    or 32^32 =7a+4^32
    therefor
    32^[(32)^32]=32^[(7a)+4^32] and here I am gone
    in the solved example, problem was easy because of 1(2^3 con 1 mod 7) but in question 2 it is not the case.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by nikhil View Post
    q 1)what is the remainder when 2^2006 is divided by 7
    ans 1) 2^3 con 1 mod 7
    or 2^2004 con 1 mod 7.......................1
    also 2^2 con -3 mod 7.........................2
    by 1 and 2
    or 2^2006 con -3 mod 7
    hence remainder is 4.
    note: con stands for congruence

    can anyone use the same procedure to find the remainder when 32^[(32)^32] is divided by 7.
    my attempt
    32 con 4 mod 7
    or 32^32 con 4^32 mod 7
    or 32^32 =7a+4^32
    therefor
    32^[(32)^32]=32^[(7a)+4^32] and here I am gone
    in the solved example, problem was easy because of 1(2^3 con 1 mod 7) but in question 2 it is not the case.
    32^{32^{32}}\equiv 4^{32^{32}}\pmod{7}

    Reduce 32^{32} mod 3

    32^{32}\equiv 2^{32}\equiv2^{0}\equiv1\pmod{3}

    32^{32^{32}}\equiv4^1\equiv4\pmod{7}
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  3. #3
    Senior Member nikhil's Avatar
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    why we reduced 32^32 mod 3 and how did it got implemented.
    what I understand (from last line)
    32^[(32)^32] con 4 mod 7= (4+7a)^(1+3b) mod 7
    how can we deduce that when (4+7a)(4+7a)^(3b) is divided by 7, remainder is 4
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by nikhil View Post
    why we reduced 32^32 mod 3 and how did it got implemented.
    what I understand (from last line)
    32^[(32)^32] con 4 mod 7= (4+7a)^(1+3b) mod 7
    how can we deduce that when (4+7a)(4+7a)^(3b) is divided by 7, remainder is 4
      4^3\equiv1\pmod{7}
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