A date conundrum
profquail: I bet they said it's wrong because they are counting "06" as "6".If you consider that each value (day/month/year) has to be at least 10, then the month must be 10 or 12 (not 11, because of the duplicate "1").Then, your day value must be somewhere in [23, 29] for October, or in [30] for December.After that, just look at the remaining 6 digits and order them in ascending order.Example: If you choose October (10), you could pick 27 as the day, which leaves [3,6] U [8, 9] for the year digits, so you'd get 10/27/3456 (US format).EDIT: In fact, I think that example may be the best answer, because choosing a day less than 27 makes the year value larger, so using 27, 28, or 29 should minimize the year. Since you want the earliest date, 10/27/3456 should be it.