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I need to find the remainder when $2222^{5555} + 5555^{2222}$ is divided by $7$. I'm thinking that Fermat's Little Theorem might help. Any suggestions?

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marked as duplicate by lab bhattacharjee, user99914, Claude Leibovici, user63181, heropup May 4 '14 at 7:12

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We see that $2222^{5555} \equiv_7 3^{5555}$. We have that $3^6 \equiv_7 1$, so $3^{5555} \equiv_7 3^5$. Similarly, $5555^{2222} \equiv_7 4^{2222} \equiv_7 4^2$. Together, we get $2222^{5555} + 5555^{2222} \equiv_7 3^5 + 4^2 \equiv_7 243 + 16 \equiv_7 0$

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  • $\begingroup$ so the answer is 0? $\endgroup$ – Jason Chen May 4 '14 at 2:56
  • $\begingroup$ Correct (and also technically any multiple of 7). $\endgroup$ – Christopher Liu May 4 '14 at 2:57

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