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Crossword Help Forum
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oyler

20th August 2016, 16:28
You have to cold solve all of the clues first then you can start to fill the grid.
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s_pugh

20th August 2016, 16:55
I must be really really stupid but I cannot understand the 2nd paragraph of the pre-amble and all the references to r and s.

This perhaps is why I cannot understand the assertion that in every pair (A/B) one must be even. Why?

If someone could enlighten this maths dunce I'd be grateful. Give me a proper crossWORD puzzle any day! ;¬)
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s_pugh

20th August 2016, 17:04
Accepting that one must be odd and one even of A/B, why must X be the even one in 9d - why not P? Really struggling to get my head around this one!
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dylan

20th August 2016, 17:06
12 ac shows that one of X or G is the even prime. If it's G, then in 9d, both components would be odd, which is not allowed, since one of them must equal 2rs
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escuan

20th August 2016, 18:34
I don't understand Oyler's comment. Since the puzzle is not a carte blanche, surely if you solve just one clue you can put it in the grid?
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pkd

20th August 2016, 18:50
Hello Escuan,

You can enter c straight away but only if the first multiple of c is too long for the grid space (eg if c was 99 and the grid space was two you could enter it because the first multiple would be 198 and too long)
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wintonian

20th August 2016, 19:45
After a lot of slogging, I've at last completed this puzzle. The ending is a bit of an anti-climax, because I don't see any interesting patterns in the final grid.

I didn't find it necessary to use Euclid's formula, though this was because I just did brute calculations on Excel. My starting point was 9d, where either P or X have to be 2. Why? well, the two numbers in the ratio can't both be even, as this would mean that the ratio isn't in its lowest terms. Also, the two numbers can't both be odd. The reason for this is that an odd number can be written as 2n+1, and its square is therefore 4n^2+4n+1. It has a remainder of 1 when divided by 4. Adding two odd squares together will give a number that has a remainder of 2 when divided by 4. Adding two odd numbers together will give an even number. However, dividing an even square by 4 will leave a remainder of 0 (an even number is 2n so its square is 4n^2). So we can't have two odd numbers in the ratio. That means that one of the numbers is odd and the other is even. This is of course consistent with Euclid's formula.

So either P or X^2 is even, but the only even prime is 2. But P cannot be 2, as no squares of non-zero positive integers differ by 4 (the squares of 0 and 2 differ by 4, but 0 is not a possible value). Hence X^2 is even, which implies that X is 2 and X^2 is 4. The only squares that differ by 4^2 = 16 are 3^2 = 9 and 5^2 = 25, so P = 3.

I was able to identify all 26 letters from 19 clues, with the other 23 clues confirming that the letters had been identified correctly. The key is to attack the clues in an efficient sequence, where you are adding at most two extra letters to those where you already have either a unique answer or at worst a very limited range of possibilities.

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unclued

20th August 2016, 20:08
I found it useful to do a Google search for "integer pythagorean triples". The first website gives you a list that I pasted into excel. For those of you struggling this really does cut down the hard work. I've just got U, F, E, L, W and Z to find.
Great news about the book and magazine oyler. How do we find out more?
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s_pugh

21st August 2016, 07:43
Thanks Unclued - that is a great piece of advice. Looks like me and Excel are about to spend some time together (again!)
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s_pugh

21st August 2016, 08:53
After a promising start I seem to have hit a brick wall as I can't get 4dn to work as PRXX:AJ+P+Q for me isn't expressed in its lowest terms. It could be that the newspaper copy has a typo (wouldn't be the first time) or more likely I've fouled up but I can't see where. Would anyone be willing to validate the 6 primes I thought I'd established so far? Thanks.
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