The only possible odd remainders from dividing number by 5 seem to be 1 and 3. The remainder of dividing by 5 in 10-based notation depends only on the last digit since every non zeroth power of 10 is divisable by 5.

From all digits from 0 to 9 only 1, 3, 6 and 8 would give the remainder of 1 or 3.

Since primes beside 2 must all be odd numbers, then all primes ending with 1 and 3 seem to be the only primes which when divided by 5 leave an odd remainder.

Hence for primes less than 100 they would be

3, 11, 13, 23, 31, 41, 43, 53, 61, 71, 73, 83