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
computer solution by Charlie)
But adding those zeros makes clear the reason for the pattern I spotted:
Each solution has a 1 digit part with no carry and a 2 digit part with no carry.
Consider 192 as two parts: 19 and 2
19>38>57
2>4>6
which together contain all the digits. The parts can be permuted so we get two solutions 192 and 219.
(zeros can be inserted between or at the end)
273 works the same way, hence 327.

Posted by Jer
on 20160413 07:21:29 