Are we doing all your homework tonight?
1)
So you want to find such that ; i.e. you want to find the inverse of 67 in the group . To do this, note that for any , ; hence , or ... can you finish now?
suppose b is congruent to a^67 (mod 91) and (a,91)=1 (gcd) for some a,b element of Z such that 0<=a<91.
1) find k element of Z such that b^k is congruent to a (mod 91)
2) let b=53. Find a
i think 1 & 2 need to be done as separate questions but I have NO clue how to do this.
any help is appreciated.