Hey PMA! Here's what I'm working on.
We are given c.
Find d and remainder e.
Using e to find starting value of a in (e,1).
In (0,1) e =2
For odd (e,1), t=1 we can calculate a var as (e+1)/2
Example: in column 3 (3,1,1) a=2
For even (e,1), t=1, a var = e/2
Example : in column 6 (6,1,1) a=3
So we can find a at t=1 starting from only c. I think so, please check my work, lads.
If true, we can now calculate x,t,b, and n for element one (t=1) in any column. Odd e and even e have different equations for t.
Knowing the pattern of how a and b drop and move as t increases in row 1, we can literally create a ladder to run t up as high as we need to to find a match for c???
VQC said all factors for a column exist in row one, the "One Row to rule them all," and if all factors exist there, then all potential c's exist there also.
So if we can ride the t elevator up to our c value, or maybe even multiple a and b parings that equal c, then maybe from that height we can look around the grid to find other locations where the answers are similar.
PMA, your method to move between (e,n) and row 1 would help us find other matching places in the grid record.
I know I've got more work to do on this, so please let me know if you find errors or have corrections. I've been feeling stuck a bit, but persistence and focus will win the day. Let's keep searching!