Possible remainders are integers from 0 through 16, inclusive: P = Q x 17 + R ------------------ 17 = 1 x 17 + 0 103 = 6 x 17 + 1 2 = 0 x 17 + 2 3 = 0 x 17 + 3 89 = 5 x 17 + 4 5 = 0 x 17 + 5 193 = 11 x 17 + 6 7 = 0 x 17 + 7 59 = 3 x 17 + 8 179 = 10 x 17 + 9 61 = 3 x 17 + 10 11 = 0 x 17 + 11 29 = 1 x 17 + 12 13 = 0 x 17 + 13 31 = 1 x 17 + 14 83 = 4 x 17 + 15 67 = 3 x 17 + 16 This might make you wonder if this is true: For any prime t and any integer r, 0 < r < t-1, there will always be a prime p such that when p is divided by t, the remainder will be r. We just proved it true for t=17. Edwin