Smithax, your across is correct, but the answer to 21d is 3-digit, so only a single digit gets inserted, not 10. As the instructions indicate, insertions cannot cross.
smithsax refers to 22d (not 21d) which does indeed have 10 inserted into the last two cells of the entry. It is not the case however that the 11 inserted into 32a has to occupy the last two cells of that entry, so the preamble can be correct.
The problem I'm having is that I cannot see how the instruction can start with A which would make 4d and 5d start with 4 and 5 respectively. This would then surely invalidate 13a which is 3755. Therefore 1a must start 145211 which means the instruction has to begin with H. Am I missing something? I await any comments to put me on the right track. If not then I capitulate gracefully.
uncle_w, you may be misunderstanding how the instruction is compiled. If 1ac has the digits 11 inserted, wherever in its length, then if A=11 the instruction begins with A. It’s the letter represented by the inserted digit(s) that matters In each clue
Just realised what I did with 32a. I put a 1 in the third cell as I knew it had to be either an insertion or the answer to the clue. The next day I added the remaining 2 ones and assumed the last two cells were the additions and not the third and fourth cells.
Too much sun I think.
Forgot to say, it's probably worth submitting your entry from abroad smithsax, enclosing a brief covering note explaining why it might be late. Even if it misses the deadline for the prize draw the nice Mr. Green will certainly allow it for your annual stats and hopefully your all correct status will be preserved ;-)
Oh well. Sigh. Sometimes wonder if the effort is worth it! When’s the next one? 12th time lucky? Or 13th? Maybe the doing is more important than the destination (a dictionary and a tiny acknowledgement in The Times one Saturday morning). : (