modular arith quetion

**yellow4321** · January 17th 2008, 10:22 AM

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**Soroban** · January 17th 2008, 11:17 AM

Hello, yellow4321!

Here's a rather primitive method . . .

Calculate: . $5^{72} \pmod{73}$

We have: . $5^9 \;=\;1,953,125 \;\equiv\;10\pmod{73}$

Then: . $(5^9)^8 \;\equiv\;10^8 \pmod{73}$

. . and we find that: . $10^8 \;\equiv\;1\pmod{73}$

Therefore: . $5^{72} \;\equiv\;1\pmod{73}$

**badgerigar** · January 17th 2008, 04:33 PM

I learnt fermat's little theorem in the form $a^{p-1} \equiv 1 \pmod {p}$ if p is prime. This applies quite easily to your problem with p = 73

**yellow4321** · January 17th 2008, 04:58 PM

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