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Math Help - Last 2 digits of 3^400

  1. #1
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    Last 2 digits of 3^400

    Hello. I want to find the last 2 digits of 3^400 using modular arithmetic. What I have so far...
    3^20 congruent to 1 (mod 25) using Euler's theorem.

    3^2 congruent to 1 (mod 4) using Euler's theorem again.

    I want to somehow reach 3^20 congruent to 1 (mod 100), and from there show that the last two digits are 01, but I have no clue. Can anyone help?
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  2. #2
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    Hello,

    Am not sure why you want to reach to 3^20

    Its quite straight forward using eulers theorem

    3 and 100 are coprime.

    So phi[100] = 100 [1/2][4/5] = 40

    So, (3^40) = 1 mod 100
    => 3^400 =1 mod 100

    So remainder is 01

    Does that make sense?
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  3. #3
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    Thanks! My textbook has hints at the back of the book, which says that I should establish that 3^20=1(mod 25), and that 3^2=1(mod4), and that finally, 3^20=1mod(100). Your solution makes much more sense . By the way, how do you calculate phi(m) in general?
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  4. #4
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by ilikecandy View Post
    Thanks! My textbook has hints at the back of the book, which says that I should establish that 3^20=1(mod 25), and that 3^2=1(mod4), and that finally, 3^20=1mod(100). Your solution makes much more sense . By the way, how do you calculate phi(m) in general?
    here's a link that may help:

    Euler Phi Function of an Integer - ProofWiki
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