find the congruence, urgent please

**knguyen2005** · May 11th 2009, 10:41 AM

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

**Media_Man** · May 11th 2009, 10:51 AM

$32^{235} = 2^{1175}=2^{196*6-1}=2^{-1}(2^{196})^6=2^{-1}(2)^6=2^5=32 (\bmod 197)$ . Using Fermat's Last Theorem, $2^{196}=2 (\bmod 197)$

**Moo** · May 11th 2009, 11:01 AM

Originally Posted by Media_Man

$32^{235} = 2^{1175}=2^{196*6-1}=2^{-1}(2^{196})^6=2^{-1}(2)^6=2^5=32 (\bmod 197)$ . 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 $2\times (-98)\equiv 1(\bmod 197) \Rightarrow 2^{-1} \equiv -98 (\bmod 197) \Rightarrow 2^{-1}\equiv 99(\bmod 197)$

**Media_Man** · May 11th 2009, 05:18 PM

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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