What is the largest number n so that n, 2n, and 3n together contain every digit from 19 exactly once?
Analytical solution preferred.
I just used excel to generate a list, then stripped off the digits and looked for a sum of 45 and variance of 7.5
n 2n 3n
192 384 576
219 438 657
273 546 819
327 654 981
What's interesting is the four solutions come as two pairs that are cycles of each other:
192 becomes 219
384 becomes 438
576 becomes 657
The second pair cycles similarly.
(The largest n is 327)

Posted by Jer
on 20160412 14:03:44 