Remainders

**Vey** · January 2nd 2010, 10:16 AM

Hey, I need to calculate the remainder when

(41^13 + 37^6) is divided with 13 and I'm kinda lost.

Thanks in advance!

**tonio** · January 2nd 2010, 11:33 AM

Originally Posted by Vey

Hey, I need to calculate the remainder when

(41^13 + 37^6) is divided with 13 and I'm kinda lost.

Thanks in advance!

Hints:

(1) For any $a\in\mathbb{Z}\,,\,\,a^p=a\!\!\!\!\pmod p\,,\,\,p$ a prime number

(2) $a^{2k}=(-a)^{2k}\!\!\!\!\pmod p$ , for any integer $k$ .

(3) $2^6=-1=12\!\!\!\!\pmod{13}$

Tonio

**Shanks** · January 2nd 2010, 11:35 AM

by the fermat theorem, the first term leaves a remainder of 2 divided by 13.
for the second term, notice that 37 leaves a remainder of -2 divided by 13, you can take it from here.

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