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Math Help - find the congruence, urgent please

  1. #1
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    find the congruence, urgent please

    Can you show me how to calculate (32)^235 (mod 197)?

    I tried to use Fermat's little theorem but it doesn't work, I dont know why.
    Thanks very much
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  2. #2
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    32^{235} = 2^{1175}=2^{196*6-1}=2^{-1}(2^{196})^6=2^{-1}(2)^6=2^5=32 (\bmod 197). Using Fermat's Last Theorem, 2^{196}=2 (\bmod 197)
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  3. #3
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    Quote Originally Posted by Media_Man View Post
    32^{235} = 2^{1175}=2^{196*6-1}=2^{-1}(2^{196})^6=2^{-1}(2)^6=2^5=32 (\bmod 197). Using Fermat's Last Theorem, 2^{196}=2 (\bmod 197)
    It's Fermat's little theorem, and it says a^p\equiv a(\bmod p), or a^{p-1}\equiv 1(\bmod p)



    So it's actually 32^{235}\equiv 2^{-1} (\bmod 197)

    Now, we know that 197=2\times 98+1
    Thus 2\times (-98)\equiv 1(\bmod 197) \Rightarrow 2^{-1} \equiv -98 (\bmod 197) \Rightarrow 2^{-1}\equiv 99(\bmod 197)
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  4. #4
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    Ah, yes. Thank you for those two corrections, cute angle. Indeed, it was Fermat's Little Theorem, and 2^{197}=2 (\bmod 197), 2^{196}=1 (\bmod 197)
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