There is an element of order 51 in the multiplicative group (Z/103Z)
Discussion: First note that (Z/103Z) has 102 elements as 103 is a prime (in fact one of the twin primes of 101, 103 pair). Also 102 = 2*3*17. So it has Sylow-3 subgroup of order 3 (prime order hence it is cyclic too) and a Sylow-17 subgroup (which is similarly cyclic). Since (Z/103Z) is abelian all it’s subgroups are normal. Thus product of Sylow-3 and Sylow-17 subgroups is a subgroup (direct product of normal subgroups is a subgroup) containing 51 elements which is again cyclic. Hence there is an element of order 51 (generator of this subgroup).