About cookies on this site Our websites require some cookies to function properly (required). In addition, other cookies may be used with your consent to analyze site usage, improve the user experience and for advertising. For more information, please review your options. By visiting our website, you agree to our processing of information as described in IBM’sprivacy statement. To provide a smooth navigation, your cookie preferences will be shared across the IBM web domains listed here.
June 2013
The way to have one digit appear twice as many times as the other nine is to set a period of 11 digits. To do that, we need the fraction to be X/99999999999. Factoring the denominator leads us to 21649 as the denominator, and looking at the possible numerators we find that 639/21649 solves the first two conditions.
To get fewer number of distinct digits, 646/21649 may seem to be the best solution (there are only two different digits), but using improper fractions, we can get the best solution: 333333/21649.
If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com