ghgonnet (07/29/82)
Cyclic Integers Puzzle. By using repeating decimal expansions we can find solutions to this puzzle. The conditions that should hold for a solution k-1 digits long are: 10**(k-1)-1 should be divisible by k and (10**(k-1)-1)/k should have no cycles in it, i.e. should not be divisible by 11,101,1001,10001,.. 111, 10101, ... and so on. As an immediate consequence, k should be prime, but this is not enough. The only solutions shorter than 100 are: k = 7, 17, 19, 23, 29, 47, 59, 61 and 97. (k=2 gives a trivial-multiple solution) The solution is the repeating decimal expansion of 1/k, 2/k, etc. E.g. 1/29 = 0.0344827586206896551724137931 034482... 2/29 = 0.0689655172413793103448275862 068965... 3/29 = 0.1034482758620689655172413793 103448... 4/29 = 0.1379310344827586206896551724 137931... 5/29 = 0.1724137931034482758620689655 172413... By noticing that if n is such a number, 9999...9-n has also the same properties, we must conclude that k*n = 9999...9; and consequently all the solutions for the puzzle are given by the above construction.