How to prove that $7^{15} + 7^{16} + 7^{17} - 1$ is divisible by $10$? This was a question on my math exam. We weren't able to use calculators so proving by manually calculating the exact value would take too long. 
In the end I ignored this question to save time but I'm still curious.
Try modular arithmetic: 
$$7^2=49\equiv-1\mod 10,$$
so $$7^{15}+7^{16}+7^{17}-1\equiv (-1)^{7}7+(-1)^{8}+(-1)^{8}7-1\equiv -7+1+7-1\equiv0\mod 10$$
and $7^{15}+7^{16}+7^{17}-1$ is divisible by $10$.
7^15(1+7+7^2)-1
=7^15(57)-1

(10-3)^15*(3(19))-1

(10*k - 3^15) *3 *19 -1
(10*K-3^16)*19-1

19*3^16-1 must be divisible by 10 if this is true

3^0=1
3^1=3
3^2=9
3^3=27
last digit of 3^4=1
               ^5=3
9
7...
repeats
-3^16 will have last digit of 1
19*-xxxxx1*-1

(10+9)*(-xxxxx1)-1
=-10*xxxx1 + 9*xxxxx1-1
=10*K  -xxxxxxx9 -1
=10*k-xxxxxxxx0
so 10 divides