Lucas27 proved the following general result relating to Exercise 37. If p is any prime number, then (n/j)
(mod p) can be found as follows: Expand n and j in base p as n = s0 + s1p + s2p2 + · · · + skpk and j = r0 + r1p + r2p2 +· · ·+rkpk, respectively. (Here k is chosen large enough to represent all numbers from 0 to n in base p using k digits.) Let s = (s0, s1, s2, . . . , sk) and r = (r0, r1, r2, . . . , rk). Then