If one tries to solve this problem without electronic assistance, one can reduce their trial and error attempts with the following information.

As the ratio of the numerator to the denominator is 1/3, our denominator needs to have the same number or one more digit than the denominator, and because there are nine digits this means the numerator will have four digits and denominator will have five. The max carry for 3

*n*, where

*n* is a digit 1 to 9, is 2, therefore the denominator must begin with a 1 or 2. And, by the same token, the smallest number the numerator can begin with is 4. (It can not begin with 3 as 3×3 equals 9, the largest decimal digit, and adding only 1 or 2 will bring the sum to only 10 or 11. There can neither be a 0 (zero) or two 1's in a zeroless pandigital common fraction of 9 digits, therefore the numerator must begin with a 4 or greater.)

For the ratio of 1/3, the denominator also needs to be a multiple of 3, therefore its digital root must be either 3, 6 or 9, and the sum of the digits are modulo 0.

Though this puzzle has already been solved, I shall still post the solution I discovered:

**Edited by Dej Mar, 25 July 2010 - 01:33 AM.**