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Hi Unclued, I decided to give it another go this morning, I'm a bit further on. But it's like walking through a bog.

From knowing O, then 4A ends in 5, also there is only one possible entry for 13A. Which gives you 6D and U and Z. Then you can enter 1D. From two possible values of I (my options were 15 or 21) only one of these can give an entry of 9xx for 1A. So now you have I, and then only 4 remaining possibilities for M due to the 4 ending on 4D.

Thanks Meursault. I'm roughly where you are but tried M=4, I=21 and K=18 to satisfy 4dn. This worked for 22ac giving 24 but then 23dn goes wrong for (LI+M)A. This is so annoying - my brain is starting to fry!

Sounds like you're on the right track Unclued, the value of M is also alphabetically appropriate. 11a is palindrome murky, and I believe the correct value for O has already been given.

I'm worried by your last comment, Unclued.

6D entry begins 5 due to value of O, and ends 4 due to only one possible 3-digit value to power O. Then I wanted to find values of Z and U. At that point I have only values 1 and 4 unallocated from 1-6. Z +10U only works for U=4.

I agree with your value for I, but I've somehow got K=20, and M either 13 or 23. This hung together quite well, since I was able to find entries for 27A and 23D which don't clash...

Could do with confirmation as to which of us has the correct values for some of these letters really.

Hi, Meursault, you are correct for U and K, and M is indeed either 13 or 23. A recent post gives a good hint for M, which will allow you to decide which it is.

Hi murky,

I think you got the right value for M, which is a prime number. You may be misreading 16ac, as the term (I+M) is multiplied by A, not divided, so it is not the case that M is divisible by A.

Just reread my last post and my logic is not precisely accurate. In 16ac, the fact that the term is (I+M)A rather than (I+M)/A means that it is not necessary for M to be divisible by A but does not make it impossible. In fact, M isn’t divisible by A, but this is not a logical implication of 16ac.

I have two options for P at this point

Thanks, wintonian. I had just discovered my error before checking the forum again. I had all the correct values for M I K when I posted my previous post. As was commented earlier, this sort of puzzle tends to fry ones brain hence reading the last bit of 16a as division not multiplication.