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Thanks Novice. At least some firm ground to start from when I resume tomorrow !

This was certainly a challenge! I made two starts on Saturday and both times, after over two hours of calculations, I ended up with contradictions. Today, after reading the forum, I realised that the key to reducing the computations was to work on clue 14. Here, the restrictions mean that there is only one possible answer, and this determines the answer to clue 2 and limits the possibilities for clue 3 considerably. Working on clues 6 and 12 together left only three possible combinations, and then working on clues 8 and 9 reduced these to 1. This gave the second, fourth and fifth digits of 16, and as this was a square and had to begin with 1, 3, 7 or 9 (given that clue 5 is a prime), there was only one possible solution.

I then solved clues 11, 7, 5, 15, 19, 38, 24, and 42 fairly directly, followed by clues 27 and 32.

I then worked on clues 13, 26 and 29, in combination with clue 36. This provides enough information (along with the third digit of clue 20 being 8 as implied by the second digit of clue 10 being 9 and 20 being 2 x 10) to restrict the possible values of clue 25 to 19881, 48841 or 77841. Further work on 26, 36 and 39 shows that only 48841 is possible.

I was then on to the digit sums. I solved the clues 26, 35, 13, 39 and 40 using the condition that the digit sum of 35 equals the digit sum of 26.

There was still a lot of tedious calculation to go, including testing thirty-six possible palindrome answers to clue 22 to find the one possibility that was a multiple of the answer to clue 2.

This puzzle was testing right to the end, but I don't think that it was particularly satisfying - no "aha" moment at the end, unfortunately.

By the way, in the statement "the digit sum of 34 = the digit sum of 16 > the digit sum of 10", I have no note of using the final condition, and looking at the answers to the relevant clues, it seems to be redundant. Is this right? As far as I can see, once the second and third digits of 10 have been determined from the second digit of 2 and the second digit of 3, whatever the first digit of 10 is, the digit sum of 10 will always be less than the digit sum of 16.

Apologies for the long post, but I'm letting off steam after what feels like a not particularly productive weekend wrestling with this puzzle.

I don't blame yoyu for letting off steam. It's a very unsatisfying puzzle, bordering on unfair, involving, as it does, scanning long lists of five-digit squares or primes. It's also possible to get quite far before things go awry, making it difficult to backtrack and find the wrong turning.

I think I have managed to establish the value of 13, and it certainly leads via 26 and 29 to a prime for 31, so I have just the SW corner to complete. Just praying I don't hit a snag down there. So far I have reached an impasse three times and had to check everything from the beginning again.

Finally got there! My last answer was 23ac and I was relieved to find that 11213 fitted the bill (I hope). I'm glad I don't send in entries because there would be a great deal of checking to be done. I have to say that I have the greatest admiration for Piccadilly for creating this thing which was much tougher than his crossWORDS which regularly appear in the EV series.

I battled with this for hours yesterday and thought I was making headway until I reached a blocker with a digit sum. I was looking at this post and was concerned at the mention of there being the same 3 digits appearing in sequence in row 3, as I cannot reconcile this. For the record I have 14ac as 118, which would suggest that 13 would have to start with a 1 and I cannot see how this can be given it has to be greater than both primes at 5dn and 7ac. Have I gone horribly wrong or are there really 3 consecutive repeated numbers in row 3?

S, 13dn starts with a 2. Your 14ac is correct.

Thanks unclued - so I take it the earlier reference to the 3 consecutive repeated numbers in row 3 is incorrect? If it isn't that would suggest my 14ac (92416) is completely wrong ....

Your 16ac is indeed correct 92416.

I haven't got three consecutive digits the sam in row 3.

s_pugh, I assume you are referring to 16ac, not 14ac. as 92416. That is correct.

I think you have misinterpreted an earlier post regarding row 3. That post by meursault commented that three consecutive numbers in row three also appear in row one (7ac). That is the case, but they are not numerically consecutive (eg 567).