$1^{2017}$+$2^{2016}$+$3^{2015}$+$4^{2014}$+$5^{2013}$+----------+$2017^{1}$.
last digit of the individual term...
$1^{2017}$ = 1
$2^{2016}$ =6
$3^{2015}$=7
. . . . . . . . .
$2017^{1}$=7
what to do next??
What will be the last digit in the following case??
1
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number-theory
elementary-number-theory
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0Last digit of $2017^1$ is $7$, not $1$. – 2017-02-02
1 Answers
0
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.