Remainder of $\dfrac {19^{81}+19^{49}+19^{25}+19^9+19}{19^3-19}$

[Albert1]

please find the remainder
$\dfrac {19^{81}+19^{49}+19^{25}+19^9+19}{19^3-19}$

 

[kaliprasad]

Spoiler

we have denominator= $19*(19^2-1)$

numerator
= $19*(19^{80} + 19^{48} + 19^{24} + 19^{8} + 1)$
= $19*((19^{80} - 1) + 1 + (19^{48} -1) + 1+ (19^{24}- 1) + 1 + (19^{8}-1) +1 + 1)$
= $19*((19^{80} - 1) + (19^{48} -1) + (19^{24}- 1) + (19^{8}-1) + 5)$
so numarator mod denominator
= 5 * 19 = 95

as $19^{2n}-1$ is divisible by $19^2 - 1$

so remainder = 95