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Ignore my post @ 10. The only solution that I have found is where (the sum of the digits in 3a) * (the sum of the digits in 6a) = the sum of the digits in 3d.

I feel like the “way in” must be through 7a, because there are only three possible second digits for one grid, which I think forces a certain second digit in the other grid, to make 3D possible. But I still can’t make it work. Am I thinking correctly (or incorrectly)?

It appears that both the clues for 9a can apply to either grid. Is this correct?

Yes captaincoma, that's what I have.

Finally limped over the line - but it didn't help being an idiot!! Whilst attempting to fill the LH grid I kept getting the 7a & 9a giving an impossible solution for the other 3d, so kept starting (and starting) over again. It was only this morning that I realised that it didn't have to because that solution is in the other b****y grid!! That is the trouble with maths puzzles, if you get fixated on a particular route and don't refresh your mind, you are undoubtedly heading for failure. I would like to change my name to twit!!

Am I being particularly thick today?
Surely the last digits of ,3d and 9a are the same,which cannot be, according to the preamble

Although the preamble says "once as an entry's last digit" that doesn't preclude it from being the last digit of more than one entry. At least, that is the way I see it.

There are 11 clues, so clearly one cell must be the final cell of two of them.

If you look at each grid, these cells must all contain full set of different digits:

Row 1: cells 1, 3, 4
Row 2: cells 1, 2, 4
Row 3: cells 1, 3, 4
Row 4: cell 1

All the other cells end an entry, and those cells must also contain full set of different digits. That’s all the preamble means

Thank you, Gitto and Quisling. That confirms my understanding. The wording. '[A]n' entry and 'once not' are a little too cryptic for my liking, but I think I see the intention of the setter.

Got there. Worth the effort! Thanks, Oyler.