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 a prime number (2) , for any integer . (3) 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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