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# Crossword Help ForumForum Rules

#### smithsax

17th November 2017, 22:34
Got in late this evening and thought for the first time ever that I might have a look at the on line Times Crossword Club version rather than waiting to buy the paper on a Saturday morning as I normally do.
Now regretting it.
Perhaps it will make more sense of it tomorrow........
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#### smartie

18th November 2017, 01:12
I got a headache just trying to understand the preamble. Reminds me why I don't bother with the mathematical puzzles. I'll be back next week!
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#### meursault

18th November 2017, 11:54
Smithsax, I found that tabulating the possible options for the 22 letters was a useful first step. Really not so difficult, 9 eligible non-zero squares, and I make it 28 different sums. Then I looked in particular at clues 12, 16 and 11, 13...
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#### dylan

18th November 2017, 14:18
After several hours detailed calculations, I've just got to T = e = 50. Anyone else found the same problem?
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#### meursault

18th November 2017, 14:45
Dylan, I have T=50 or 58
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#### dylan

18th November 2017, 15:38
Oh dear, I got N=58!

The problem with these puzzles is that it's very difficult back-tracking to find an error- and I don't think I have the will-poeer to start again from scratch!
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#### meursault

18th November 2017, 16:35
I'm not saying you're wrong Dylan. But assume you start by entering 16. Then, laboriously, eliminate all the possibilities for 12(2) apart from two. The entry for 16 excludes one of those. So, then there was only one possibility for 12(3), and that ends in 9. So, 14 must be 9x and is the sum of 2 squares, the possible values for x being 0,7,8. You already have the first digit for T entered, the possible values for the other are 0,2,3,8. Only 0&8 correspond.

However I may easily have made an error since none of it is self-checking, at least not yet. And values of either 50 or 58 for T are giving me a problem with the value for M in the second expression of 14.

I need to get out !
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#### demeter

18th November 2017, 16:40
I've solved the algebraic part and got the three-part hint, but still can't work out what to do.

Dylan,

I initially made the same mistake. I think what you've probably done is forgotten to allow that both the single expressions, and the 2nd and 3rd expressions in the 3 expression clues, may be negative. This means that there are a few more branches to consider (which you presumably disregarded en route to arriving at two numbers occupying the 50 spot).
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#### meursault

18th November 2017, 17:03
I hadn't considered negative values either, Demeter. Thanks for the warning.
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#### captaincoma

18th November 2017, 19:39
I was going along nicely but ended up with E=a=53 and Y=N=65. Can someone please tell me which is correct?
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