What is the unique three-digit positive integer satisfying 100x = 11 (mod 997)
11 + 997a = 100x
Since 11 + 997a needs to be divisable by 100, the last 2 digits need to be 00.
11 + 997a = 0 (mod 100)
11 + 97a = 0 (mod 100)
a = 37
11 + 997(37) = 100x
36900 = 100x
x = 369
=^._.^=
