Hey, I need to calculate the remainder when (41^13 + 37^6) is divided with 13 and I'm kinda lost. Thanks in advance!
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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 $\displaystyle a\in\mathbb{Z}\,,\,\,a^p=a\!\!\!\!\pmod p\,,\,\,p$ a prime number (2) $\displaystyle a^{2k}=(-a)^{2k}\!\!\!\!\pmod p$ , for any integer $\displaystyle k$. (3) $\displaystyle 2^6=-1=12\!\!\!\!\pmod{13}$ Tonio
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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