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Oops, this is an immediate reply to my own post. I think I now understand my error. I was for some reason assuming pentominoes couldn't cross the grid's internal walls, but I now think that is wrong. Without that restriction, I can see how the necessary pentomino might be constructed.

I have finally cracked this. I could not agree more with catkin@22 and murky@35 which is why I thought some cells must have more than 1 digit in. With thanks to quisling's lucid explanation @23 I then realised what I needed to do. I do find it frustrating when the preamable actually turns out to be more difficult than the crossword? itself.

Would anybody out there have a copy of all THE CLUES (not available on the Listener website) to LC 4000 ?! With the mini-milestone of LC 4700 coming up, I'd like to have a crack (in anniversary mood) at this puzzle! I do have the GIANT grid but not, as I say, the clues .... anything, you might say, to distract from these now very dark days (in addition to Covid, as if that wasn't enough). Any help / hints will be most welcome.

Primeprospect, if you drop me a line at patmark291 at yahoodotcom, I should be able to help you

Pp – I have the clues for Listener 4000 that I evidently downloaded from Dropbox way back then (I’ve been meaning to have a crack at it for years…) - it’s a huge 38 Mb PDF file (why, I’ve no idea – it’s only a few pages of b&w text or images, after all) – but my laptop is struggling to display it – and it will almost certainly not transmit without problems, unless I can find a way to shrink it radically by (eg) OCR?copy-and-paste into Word – or some such.
Then, Norah, for very good security reasons, doesn’t want us to display e-mail addresses in the Forum, so I’d have to find some other way to get it to you – this chat window accepts URLs, so maybe I could put the smaller file back on Dropbox – if that doesn’t breach Copyright - but I don’t know how to do that.
All in all, if you get a better offer, go for it!

https://pdfcompressor.com/

You drag your file to the box and it does its best. Works beautifully on some pdfs and hopelessly on others but well worth a try

https://www.ilovepdf.com/compress_pdf is the other one I have used with good results

Jordan - thanks; I'll have a go.

Well, here I am, feeling more stupid than usual, still hung up on 4699.
In the hope that someone is still watching this thread (everyone else seems to have found it easy) here's my situation:
I've worked out what number 7a has to be, and where it must go to form any sort of pentominoes, and one pentomino that as far as I can see has no alternative, to contain 5 of its majority digit. I've then used this filled pentomino configuration to deduce 3d and 4d - although which goes where, L or R, is not clear. I've also deduced a plausible value for 5d.
My problem is that the first step - entering 7a and completing the only possible pentomino filled with its majority digit - leaves its minority digits in positions which can't form unique pentominoes, while also satisfying the requirement that the top two-digit down entry is a prime.

To put it slightly differently: the first minority digit in 7a cannot form a pentomino which includes the second digit of the two-digit down entry at top centre, as this entry is clued as a prime. The only possible alternative filled pentomino configuration for the first minority digit in 7a then prevents the possibility of the second minority digit forming part of a unique pentomino - as far as I can see

I'm not sure how anyone can help me sort this out without giving too much away - maybe I've given too much away here myself - but if anyone can deduce my position for 7a from the above, just a "correct" or "wrong" would be helpful...I hope...

(In the meantime I've sorted and submitted 4700!)

Hi Quixote,

If you have worked out 3dn and 4dn, and have a good idea of 5dn, you must know what prime number is represented by G. If you don't, you can deduce that G must be a single-digit prime, as 5dn has three digits and so do 3dn and 4dn, which are basically 5dn multiplied by G. If G had two digits, then 3dn and 4dn would have four digits.

You already know two of the four single-digit primes from 7ac (C and H). You can eliminate one of the other values easily, because GKL+/-C will either contain a zero, which is not possible as the grid contains only the digits 1 to 9, or either K or L will have to equal G.

This means that there is only one possible prime value for G, and as 6ac is ACDG, you can work out what the final digit of 6ac is - you should find that it is the same as the second digit of 7ac. Given the placement of 7ac in the grid, the answer to 6ac can go in only one position, and this limits the possibilities for the pentomino containing the second digit of 7ac. As you correctly note, the second digit of the two-digit entry at the top of the grid cannot be the same as the second digit of 7ac, or the entry would not be prime.

I hope that this gets you to the next stage. I found that there was only one possibility for the pentomino containing the second digit of 7ac and this meant that there was only one possibility for the pentomino containing the fourth digit of 7ac. If you can work out the shape of 6ac, in the only position it can occupy, you have just enough information, I think, to divide the whole grid into nine distinct pentominoes, and the other numbers fall into place quite quickly.