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agreed - the puzzle is tough enough without red herrings being introduced.

I agree. I've finished, got the cipher and the thematic word but haven't a clue what some of these comments refer to. After trying some more complex ciphers, the actual one was the simplest of the lot.

There are several completely systematic ways of encoding, each giving a different. One really need to guess the word and work from that. Crosswhit's comment on anagrams suggests I have the right one, but that can be arrived at by two different sensible routes, neither of which gives me any coherent words in the grid except the odd three-letter word. There seems to be a scarcity of vowels and a concentration of consonants. I can see 16 letters that probably remain in the grid. The rest I'm unsure of.

I don't think I've ever been quite so bored by a puzzle. Time to call it a day, methinks, though having waded through the mire thus far I'm reluctant to give up. I'll pray for inspiration overnight.

Murky, I sympathise. Probably I have been as bored previously, though that time would also have been a numeric puzzle.

No don't give up, but I do recommend a change of approach. The thematic word can be left until last. Then it appears quite easily.

I think that you're not sure what numbers to move into Grid B. When I was trying to decide this, I noticed some similarities in Grid A rows 2-5, in fact 4 columns within those rows are identical. Then I could see that there were some common features concerning the columns (in rows 2-5) to the right of each of those 4 columns. Except that there is a strange anomaly at the bottom right. Don't worry about that. It took me a little while then to start reading all these cells from left to right, rather than vertically. Then it became clear to me which 32 cells should be copied, leaving only a central line and perimeter to be filled in Grid B.

Now you need a cipher. In fact you already have one, which you've created by writing letters against numbers when you solved Grid A. See if the remaining 34 numbers from Grid A make any sensible words using the possibilities from this key. No ? I thought not...but now say to yourself, what similar key could I use ? Don't over-complicate it, just think of the most basic key along similar lines to the one which just failed (26 letters, 26 numbers).

Hi all. I’m relatively new to the listener having been solving it for about 6months (or attempting to) often with help from this site. I’ve come in rather late on this one and finding it very difficult. I’m confused by the fact that the expressions containing equals signs eg 11A and 21D do not seem to be equivalent ie if you substitute numbers for the stated letters and solve the equation the two halves do not seem to equate! I have not as yet determined the correct numbers but the equation should still hold with any substituted number. Or am I totally barking up the the wrong tree!

Goshawk, the expressions either side of the equals sign do give the same answer, but only if you substitute the correct numbers for the letters. That is how you set about solving a puzzle like this.

Thanks for your comments, Meursault. They could prove very helpful in getting over the last hurdle. I had noticed all those features in grid A to which you have drawn attention, though I was puzzled by the two anomalies in the SE corner.

Ok. Thanks murky. My maths is a bit rusty! It’s only the second numerical I’ve attempted. Will proceed with help from some of the previous posts.

Hi, goshawk,

It's best to think of the various clues with equals signs as giving different algebraic expressions for the same answer. For example, in 21dn, the answer can be expressed as HO-T/E-L or as HO-TE+L. You can use the fact that these two expressions lead to the same answer if you substitute the correct values for H, O, T, E and L to work towards these values.

As HO-T/E-L=HO-TE+L, we can ignore the HO term on both sides of the equation and we have -T/E-L=-TE+L. Rearranging this gives TE-T/E=2L. As T has to be a multiple of E (if it isn't, then TE-T/E can't be a whole number, which is necessary as the right hand side of the equation is 2L), and E, L and T are all whole numbers between 1 and 26, the maximum value for 2L is 52, and it turns out that there are only 13 possible pairs of E and T that give 2L less than or equal to 52, with L a whole number.

Similarly, in 11ac, we can treat the two expressions as an algebraic equation, and divide both sides by A (which is legitimate as A cannot be zero) to give D(E+L+T)=D+(E+L)T. I then set up a spreadsheet to calculate the two sides of the equation for the 13 combinations of E, L and T that I had from 21dn and all possible values of D, and found that there was only one value of D and one combination of E, L and T for which the two sides of this equation were equal. This is sometimes called "brute computation", because it involves a lot of calculations. However, D must be a factor of (E+L)T, and using this fact will reduce the number of calculations considerably.

The point is that the equations will not hold with ANY substituted numbers, only with a limited set of combinations, and in this case there is only one combination of D, E, L and T that satisfies the equations in both 11ac and 21dn.

The positioning of the five zeros in Grid A is also significant murky, as after enciphering there is one per word in Grid B.