+0

+1
338
1
+567

Determine the number of positive integers a less than 12 such that the congruence $$ax\equiv 1\pmod{12}$$ has a solution in x.

ant101  Apr 2, 2017
#1
0

By simple inspection, we have:

a=1   and  x=1

a=5   and  x=5

a=7  and  x=7

a=11 and x=11

Guest Apr 2, 2017