[VA] Short & Sweet Math Challenges #23: "May the 4th Be With You !" Special

[Albert Chan]

(11-04-2019 07:18 PM)Jeff O. Wrote:  My logic for this was as follows:

1. Let the digits^n of n-digit number A sum to n-digit number B
2. Let the digits^n of n-digit number B sum to n-digit number C
3. If B=C, then the digits^n of n-digit number C must also sum to n-digit number B
4. Therefore, number C (and B, of course) must contain the same digits as number A, i.e., is a permutation of the digits in A.

No. 4 is a conjecture, not an assertion ...

Hi, Jeff O.

I checked all (88) Armstrong numbers, and see how many A's can produce it.

If B is an Armstrong number (B=C), A is unique → A,B has same sets of digits

Confirmed (by brute force) that your conjecture is correct.