Find the smallest distinct positive integers, M and N such that you can rearrange the digits of M to get N, and you can rearrange the digits of M
3 to get N
3. [Leading zeroes are not allowed.]
102^3 = 1061208
201^3 = 8120601
This was based on the observation 12^2 and 21^2 are reversals as are 13^2 and 31^2. The extra 0 compensates for carries.
I don't know if there is a two digit solution though.
|
|
Posted by Jer
on 2024-03-22 09:10:46 |
A guess
