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......had another look....... and now get 2 possible answers for 3d and 10a which still fit with the other answers...... same 2 digits for 3d but ending in 4 or 6....same 2 digits for 10a but ending in 4 or 6 ???How can one pair be eliminated?

No two grid entries are the same.

Of course! Many thanks - it's the 4's then!!

There must be a flaw in my logic, but I can't see it. Any help gratefully accepted: 4d/19d must be 50, 52, or 70. If 19d ends in a zero, so does 21a which must give 370 but this is impossible since 15d requires the middle digit to be 0, 2, 4, 6, or 8. So 19d is 52 and 4d can't be (no two entries the same) so 4d ends in 0 and the middle digit of 8a is 0, which is impossible since the second part of the entry can't start with 0 and the first part must be prime...

370 is not a possible grid entry for 21a.

Yes, I realised that (see my earlier posting), but I had overlooked that a final zero in 21a does not imply a factor difference of exactly zero, the difference could be a multiple of 10, with a lower factor of 9 or less, so my problem is resolved.

I have 760 for 21Ac.

Carpox1, what I meant by that is that 370 is incompatible with the 12(3,3) specification for 21A. Only two numbers fit this.

Apologies, demeter, if my response seemed ungracious (and ungrateful): I take your point. Thanks, unclued, I too have 760 (now!) - the structure of which was the originally overlooked possibility to which I alluded (the final zero stemming from a two-digit multiple of ten).

No problem Carpox1. Glad you've finished.