$1^{2017}$+$2^{2016}$+$3^{2015}$+$4^{2014}$+$5^{2013}$+----------+$2017^{1}$.
last digit of the individual term...
$1^{2017}$ = 1
$2^{2016}$ =6
$3^{2015}$=7
. . . . . . . . .
$2017^{1}$=7
what to do next??
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Last digit of $2017^1$ is $7$, not $1$. $\endgroup$ – barak manos Feb 2 '17 at 7:43
Add a comment
|
$\begingroup$
$\endgroup$
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.
