Tripartite Palindromic Partition of Integer (HP 50g) Challenge

[2old2randr]

(03-10-2023 11:43 AM)3298 Wrote:  While I haven't looked at the UserRPL program, from the proof alone I can see that it runs into one of the mentioned choose-numbers cases in the 6-digit algorithm. There it apparently disregards the non-zero digit restriction and chooses \(x_1=9\) and \(y_1=0\), causing the y palindrome to become invalid.

That is exactly correct - I've handled it correctly in other positions but not for x1 and y1. I'll fix it on the weekend.

Sudhir