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$1^{2017}$+$2^{2016}$+$3^{2015}$+$4^{2014}$+$5^{2013}$+----------+$2017^{1}$.

last digit of the individual term...

$1^{2017}$ = 1

$2^{2016}$ =6

$3^{2015}$=7

. . . . . . . . .

$2017^{1}$=7

what to do next??

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  • $\begingroup$ Last digit of $2017^1$ is $7$, not $1$. $\endgroup$ – barak manos Feb 2 '17 at 7:43
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All of the bases can be reduced mod $10$. All of the exponents can be reduced mod $\phi(10)=4$. So the sum any 20 consecutive terms should be the same.

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