find remainders when 2^50 and 41^65 are divided by 7. this is a mod problem but I don't know how to work it...

Answer 1

By Fermat's little theorem we have:

a^6 = 1

(equality means equality mod 7)

if GD(a,7) = 1.

2^48 is thus equal to 1, therefore

2^50 = 2^2 = 4

To compute 41^65, we can use that
41 = -1, and taking the 65th power then gives:

(-1)^65 = -1 = 6