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Thanks Ollie - penny just dropped

There is only one place the 0 can go, once 20a is solved. Now have 20a, 1a, 3d, 17d, 18d and partial,answers to 16a and 12d. Can't see how to work
out the other answers. Hint appreciated.

You need to look at 16a and 5d in tandem and see which possible combinations give you a viable set of coded numbers in conjunction with completing the top and bottom rows.

Thanks again cruciverbalist. Unfortunately I seem to have made an error somewhere. Maybe now is the time to give up and wait for tomorrow's offering.

I'm really stuck on this one and haven't made any progress since Friday. If I list my reasoning so far, perhaps someone could point out the flaws in it.

First, there seems to be only one possible answer for 20a, and only one possible grid entry for 1a that ensures both entries are squares and all entries in the grid are unique.

Then, obviously we have the clue answers for 3d and 16d, and it is clear that 0 is the second digit of 19a. By multiplying all numbers of the form _3 by _8, to find the answer for 20a, there seems to be a unique solution for 17d (and clue answer for 20a), determinable without filling in any more of the puzzle.

Because 12d is the square of 1d, its last digit must be either 0 or 1, and since 16a is a multiple of 5d, I determined it must be 0. At this point I have the (apparently incorrect) permutation of all digits.

Now here's the problem; the last digit of the answer for the 9d clue answer is clearly 0 and it follows that the first digit of the answer for 7d is 0 doubly encrypted, which completes the grid entry for 6a, giving an answer which is not 12d plus a square.

I'm sorry if this is a bit vague. I didn't want to give too much away.

I’d like to help, xmogg, but I’m not even sure about 20a. My reasoning, such as it is, is this: suppose the entry for 20a (which is a square) is ABC, then if the entry for 1a (which is also a square) is DEF, we know that A encodes to D, B to E, and C to F. We also know that A, B, C, D, E, and F are all different. The clue to 3d is the square root of ABC, which is, let us say, GH encodes to FI because of the overlap of 1a and 3d. The middle digit of the entry for 20a, B, must be the encoding of an even number (from the clue to 15d). The final digit of the entry for 20a, C, is also the final digit of the entry for 18d, which is JC and we know that C encodes to J and J to C. However, I can’t see the additional constraint(s) that enable(s) a breakthrough to be made. Consideration of the squares leads me to suppose that it’s these clues/entries that permit a start (and a number of other helpful postings say as much), but I’m firmly stuck.

I'm sorry to hear that Tina - if you want to drop me another e-mail with your grid fill to date I'm happy to tell you where the error is.

(That offer is open to all btw:

cruciverbalist@btinternet.com)

"since 16a is a multiple of 5d, I determined it must be 0"

This is where your logic breaks down xmogg - try the alternative instead.

"The final digit of the entry for 20a, C, is also the final digit of the entry for 18d, which is JC and we know that C encodes to J and J to C"

In fact Carpox1, we know that C encodes to F (and therefore from 18d F encodes to C). This means that the last digit of the grid entry at 20a is the same as the first digit of its square root. There is only one 3-digit square with non-repeating digits that fulfils this requirement.

Thank you, cruciverbalist, that's just what I needed to get started. I'm most grateful.