I don't understand how to use Fermat's little theorem to find remainders e.g if we are asked to find remainder of 50^50 on division by 13, what is a and what is p in the formula? What is going on?? Are we using congruence classes when using Fermat's little theorem? Can someone please give me a step by step explanation from the beginning, I'm just not understanding this stuff.
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1$50 = 4\times 12 +2$, do you think this will help? How might you then simplify $50^{50}$? Bearing in mind that $12 = 13-1$ and $(50,13)=1$... – ah11950 Apr 24 '14 at 22:39
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Finding the remainder $r$ is the same as solving $$50^{50}=r\mod13$$ – user137794 Apr 24 '14 at 22:39
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Please do not reask your prior question. If the answer does not suffice, ask for help in comments, and/or edit your question to clarify. – Bill Dubuque Apr 25 '14 at 00:29
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$50^{50} = (50^{12})^4\cdot 50^2$. Now apply Fermat little theorem: $50^{12} \equiv 1 \pmod {13}$. So $(50^{12})^4 \equiv 1^4 \equiv 1 \pmod {13}$, and $50^2 \equiv 4 \pmod {13}$. So: $50^{50} \equiv 4 \pmod {13}$
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