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xij, i is 96 and I is 653. Keep going because this puzzle is something special.

Forget that. I just found out that I had missed out two of the cubed numbers, both similar end results.

Hi, I'm not good with numerical puzzles but I like to give them a go. A hint on where to start would be appreciated. I have the range of values for U but where do I go from there? Thank you in advance.

Supersub, there's only one possible value for U so that U cubed is a three-digit number (Through 4). That gives you a starting digit for 12a, so you can narrow possibilities for C. Look at clues with p in them next.

Make sure you have a complete list of integers. I've only just realized that I had overlooked one, making it impossible to get any equation involving r to fit my almost full grid.

You should be working with a list of 42 integers from which to select the 27 letter values, up to the cube of 46, less its leading 9 (the cubes of 1,8 10 and 20 can't be used). If you rank the remainders in ascending order (cubes of 4, 3, 8, 6, 5 etc. up to 31) the five words will soon become fairly obvious.
After establishing U, consideration of 12a (4_ _ _ 3) gives C and then 17d and 2t give p and E ......

Also the first word gives a lot of help. It’s 8 letters with two z and describes what we all are doing!

I am so impressed by the construction of the final cube. So clever.
Thanks to Elap for a delightful end game after a challenging but possible puzzle.

Magnificent, brilliant. Took me the best part of 24 hours to finish but it was worth it. I initially made 3 mistakes. Firstly I didn’t use 16 cubed because it started with 0, then I stupidly missed out 22 and 37 cubed, then, again stupidly, didn’t do 30 and 40 cubed thinking they’d be like 10 and 20. So, my message to any newcomers is “Read the preamble carefully”.
Incidentally, I cheated for the last stage. Once I’d got the first word and last three words I made an anagram of the remaining letters to get the key word. This made finishing the numeric bit a lot easier.

Why is E determined? I have two possibilities

If E were 331 (and 17d 315) then 8through would be 285 with z = 43. 2through restricts L to 43 maximum but this combination of z and L rules out 1 as a possible middle digit for 14through, and therefore 17d, so E cannot be 331.