Find an integer x such that 0 <= x < 527 and x^37 = 490 (mod 527).
I don't think that there is an interger x that satisfies x^37=490. The closest integer would probably be 1. Hope this helps!!
x^37 mod 527=490, solve for x
a=1; c=a^37 % 527; if(c==490, goto3, goto4);printc, a; a++;if(a<1000, goto1, 0)
x =527c + 56, where c =0, 1, 2, 3........etc.
The smallest x = 56, and 56^37 mod 527 =490
