How can I prove that $36^{36}+41^{41}$ can be divided by $77$? I think it is something about the rest of $36^{36}/77$ and $41^{41}/77$. But how can I find it?
Proving that a sum can be divided by 77
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elementary-number-theory
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0Hint: Modulo $77$ we have $41\equiv -36$. This leaves you with the easier task of calculating the remainder of $36^5$ (or $41^5$) modulo $77$. – 2017-01-13
1 Answers
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$$36^{36}+41^{41}\equiv 36^{36} + (-36)^{41}\equiv 36^{36} - 36^{41}\equiv 36^{36}(1 - 36^5) \pmod{77}$$ So, it is enough to prove $77 | 36^5 - 1$. Since $77 = 7\cdot11$, it is enough to prove 7 and 11 both divides $36^5-1$.
Clearly $36\equiv 1\pmod7$. So, $$36^5\equiv 1^5 \equiv 1 \pmod 7$$ Thus, $7 | 36^5-1$.
Moreover, $36\equiv 3\pmod{11}$. So, $$36^5\equiv 3^5 \equiv 243 \equiv 1 \pmod{11}$$ Thus, $11 | 36^5-1$. Done.