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Thread: finding the remainder

  1. May 1st 2009, 01:25 AM #1
    adhyeta
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    Exclamation finding the remainder

    how do we do such problems?

    like...

    find the remainder when 3^{37} is divided by 79...

    or when 4^{101} is divided by 101...

    whats the method?
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  2. May 1st 2009, 02:33 AM #2
    ADARSH
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    Quote Originally Posted by adhyeta View Post
    how do we do such problems?

    like...

    find the remainder when 3^{37} is divided by 79...

    or when 4^{101} is divided by 101...

    whats the method?

    Easy way to solve such problem is like

    since 3^4=81 we have to take value greater than 79 hence (3^4)^9*3 = 3^37

    3^4/79 = 81/79 leaves remainder 2 hence 2^9 *3 / 79 = 512*3 = 1536/79 will leaves a reminder 35

    I hope it is Easy solution (think of binomial expansion)
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  3. May 1st 2009, 02:55 AM #3
    adhyeta
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    Quote Originally Posted by ADARSH View Post
    hence (3^4)^9*3 = 3^37
    dint get this
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  4. May 1st 2009, 03:16 AM #4
    ADARSH
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    Quote Originally Posted by adhyeta View Post
    dint get this
    {(3^4)^9}*3 = 3^{37}
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