Wilson's puzzle: Given a base b and initial nonnegative integer seed x, define k(b,x) to be the smallest value k such that the base-b representations of (1*x, 2*x, 3*x, ..., k*x) use all of the b digits. Define k(b) to be the largest value of k for all x. The following are true. k(p), for p prime, is p. (A seed value of 1 suffices, and, ignoring multiples of p and just looking at the least significant digit, no other value will generate a higher result.) k(b) <= b(b-1). (Look at the digit place one greater than the least significant digit of the best seed 'x'; by the time b(b-1)*x is reached, this digit has progressed through 1 all the way to b-1. The zero is covered at the time we multiplied x by b.) For all values of nonnegative t, 1<t<b and positive n where b^n/t is an integer, we have k(b,b^n/t)=t*(b-1) (1) In order for b^n/t to be an integer, the set of prime divisors of t must be a (strict or non-strict) subset of the set of prime divisors of b. The highest k for all allowed values of t occurs for the greatest such value of t; it is easy to show such a value of t must be >= sqrt(b). We can show (1) by considering the base-b expansion of 1/t. The last digit encountered will always be b-1. We perform long division of b^n by t in base b. At every point our remainder r<t, and the next digit is floor(rb/t). We can prove that rb/t < b-1 as follows, using the fact that r<t<b, and t and b are nonnegative integers: rb/t < b-1 rb < (b-1)t t < (b-r)b Now, b>r so b-r >=1, and thus (b-r)b >= b, and b > t, so we are done. Consider next all values of i from 0..t-1. For all such values of i when computing the decimal expansion of i/t, we will never encounter the digit b-1. But at the value of i=t, we calculate 1.0, so t..2t-1 will generate the same digits as 0..t-1, except maybe also including a 1. The first time we will encounter digit b-1 is at i=(b-1)t, so k(b,x)=t*(b-1) Empirically, it appears that there is no higher seed possible; that is, that k(b)=k(b,x). I have examined values of b up to 1600 without finding a counterexample. I believe this can be proved but have not yet been able to prove it. If it is true, then for non-prime base b, the limit on k(b) is (b-1)(b-2) (t and b must have at least one common factor, and thus they must not be adjacent, so t<=b-2), and this is attained for 2+2^n for n>0. Interestingly, the case we started with, b=10 or decimal, is one of those values at the limit.