Given a = 1 (mod 7), b = 2 (mod 7), and c = 6 (mod 7), what is the remainder when a^{81}*b^{91}*c^2 is divided by 7?
a==1, b==2, c==6
[1^81 x 2^91 x 6^2]==[89,131,682,828,547,379,792,736,944,128] mod 7 ==2 - the remainder.
+506 Nice, guest, just dropping a non-computation-heavy solution
If a is 1 mod 7, then raising it to any power will still be 1 mod 7, meaning that a^81 is 1 mod 7.
For b, notice that 2^3 is one more than 7, so b^x will be 1 mod 7 for x being multiples of 3. Since 91 is 1 more than a multiple of 3, it will multiply another 2, meaning that b^91 is 2 mod 7.
Lastly, since c is -1 mod 7, c^2 is (-1)*(-1) = 1 mod 7.
Lastly, just multiply all the results together to get 1*2*1=2
