What is the largest number n so that n, 2n, and 3n together contain every digit from 19 exactly once?
Analytical solution preferred.
(In reply to
Analytic Answer by Brian Smith)
Assume n starts with 3:
Then n, 2n, 3n are 3xx, 6xx, and 9xx. There can be no carry from the tens to the hundreds in this case. This implies the tens digit in 9xx is at least 6. To get that high the tens digit in 3xx must be 2, which leaves 7 as the only possible tens digit in 9xx. Then the numbers are 32x, 6xx, 97x. This leaves only n=324, 325, or 326 as possible candidates. None of those satisfy the problem.

Posted by Jer
on 20160412 14:11:58 