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Wintonian - thank you very much for this thorough exposition; I'd already followed the path that you describe up until the solution of 6ac - but I then have the following difficulty:

At post #5 on this thread Xwordfan posted '6a in top left corner' - if this is correct (and as you say 'the answer to 6ac can go in only one position'); and if, also as you say '...the final digit of 6ac is......the same as the second digit of 7ac' , I cannot for the life of me see how any pentomino connecting the last digit in 6ac with the (identical) second digit in 7ac can possibly avoid putting the same digit in both first and second places in the two-digit down entry at top centre - which will then not be prime - so, wrong.
As you say:' 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.'

The only explanation for this paradox that I can think of is that I've put 7ac in the wrong place - but I really can't see any alternative, and 3dn and 4dn seem to confirm its position.

As you and everyone else have solved this, I must be being extremely dense.

Any further comment? I've missed the post for submission, but don't want to just give up.

Wintonian - thank you very much for this thorough exposition; I'd already followed the path that you describe up until the solution of 6ac - but I then have the following difficulty:

At post #5 on this thread Xwordfan posted '6a in top left corner' - if this is correct (and as you say 'the answer to 6ac can go in only one position); and if, also as you say '...the final digit of 6ac is......the same as the second digit of 7ac' , I cannot for the life of me see how any pentomino connecting the last digit in 6ac with the (identical) second digit in 7ac can possibly avoid putting the same digit in both first and second places in the two-digit down entry at top centre - which will then not be prime - so; wrong.
As you say:' 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.'

The only explanation for this paradox that I can think of is that I've put 7ac in the wrong place - but I really can't see any alternative, and 3dn and 4dn seem to confirm it.

As you and everyone else have solved this, I must be being extremely dense.

Any further comment? I've missed the post for submission, but don't want to just give up.

Oops -sorry for the duplication - I'd forgotten that there can be a delay , and re-posted thinking I'd failed to pass the Captcha

Aaaaahhh - the light may be dawning - I let myself be intimidated by that vertical black bar.....

..gee that's bettah
Muddah, Faddah kindly disregard that letter.

Let's see where this goes

Nailed it!...finally.

Thanks to Wintonian - and not least to https://metanumbers.com/

Hi Quixote,

Just caught up with your earlier post and glad that you have sorted things out. Yes, pentominoes can (and do) cross the bar lines.

Because of the pentomino restriction, there are only two possibilities for 6ac that have the right shape to fit in the first across entry (of four digits). One of these must be rejected because it would require the second digit of the upper 2-digit down prime entry to be even. The remaining entry uses two of the four digits from 1, 3, 7 and 9. Another of these is used in 7ac, so the lower two-digit prime down entry has to end with the remaining number. Because you have already used a particular shape of pentomino for the repeated digit in 7ac, you are restricted to a specific pentomino that includes the second digit of the lower 2-digit prime number and also the second and third digits of the lower down entry in the extreme right column. You should notice that the lower down entry in the extreme right column turns out to be 5dn.