Prove that
$\displaystyle 1^{1987} + 2^{1987} + ... + 13^{1987}$
is divisible by $\displaystyle 15$.
Hello,
I guess there is a typo.. (in red)
$\displaystyle 14 \equiv -1 (\bmod 15)$
$\displaystyle 13 \equiv -2 (\bmod 15)$
$\displaystyle 12 \equiv -3 (\bmod 15)$
$\displaystyle 11 \equiv -4 (\bmod 15)$
etc...
So $\displaystyle N=1^{1987} + 2^{1987} + ... + 13^{1987}+14^{1987} \equiv 1^{1987} + 2^{1987}+3^{1987}$$\displaystyle +4^{1987}+5^{1987}+6^{1987}+7^{1987}-7^{1987}-6^{1987}-5^{1987}-4^{1987}-3^{1987}-2^{1987} -1^{1987} (\bmod 15)$
$\displaystyle N \equiv 0 (\bmod 15)$