Suppose that (a, m) = 1. if a = +1, the solution of ax = 1 (mod m') is obviously x = a(mod m'). if a = +2, then m is odd and x = (1 - m')}a (mod m') is the solution of ax = 1(mod m'). for all other a use problem 11 to show that the solution of ax = 1(mod m ) is x = k (mod m') where k is the nearest integer to -(1/a)(1 – ax,)'.