Rich wrote: << Q: What's the reciprocal of 33 mod 100? Hint: Add 64. When N ends in 3, the (mod 100) reciprocal of N is N+64. So the reciprocal of 33 is 97. 33*97 = 3201 = 1 (mod 100). The whole algorithm ... Suppose the tens digit of N is T. If N = 10T+1, then 1/N = 102-N 10T+3, N+64 10T+7, N+36 10T+9, 98-N
So I wondered if this kind of thing always happens for the square of a product of two primes, and programmed it up for 441 = (3*7)^2. But my eye can't pick out any pattern as simple as this one for 100. Hmm, does something like Rich's observation always work if one of the primes is 2, i.e., for numbers of the form 4 p^2, p prime ? --Dan I sleep as fast as possible so I can get more rest in the same amount of time.