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To sum up, gitto and andyp, have three creators who fit at 1ac, can see (different) forenames of two of them in the grid. All three need a '7letter' to be used to 29 the unnumbered.

Make that three forenames!

greengage - sounds like you now have what you need to 29

Thanks andyp, let's hope so!

I now have three forenames, the last I found was the creator of the technique that the mathematician solved, That now means I still do not know what to do because I'm thick.

you therefore should have three "variants of 7" (that comply with the preamble). one of these is to be used "inside" the other two (applied successively) to 29 the unnumbered entry.

I will readmit here that i didnt work out the 29'ing process without considerable help - i managed to GWIT to the final result with help from the preamble, and then the help came afterwards.

andyp - not sure what "inside" means. I have at last identified 3 chaps with different methods of 7, and I assume I use each one once. The preamble says that at all stages there are real words (apart from the initial stage in the filled grid). Does that really mean that after each of the 3 29ing processes we have real words? Can we ignore the Latin chap in this set of 3 29ings?

sorry - trying not to give it way so being perhaps a little obtuse. you only use two of the methods, but the third os still critical (think what you need for each of the methods)

Mazel tov to those who have got this over the line, but this is clearly a GWIT puzzle, however clever. I have found all three names, I have all three methods, I've got no idea what I should be using for a "7" from the long message and I see no way of distinguish which method is first, second or third. It's puzzles like these that make me think about giving up on the Listener and getting my weekends back.

I'm with you twelvetrees. It's too much.
I have full grid, I marvelled at the messages yesterday. But now finding the seven letter K?????? is a step too far.
I'm happy to let it go.