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The rubric does not say the 16 throws are unique or that W, X, Y and Z cannot be 1. The three squares can be quickly deduced as can the value of X. The rest follows but I found it a bit of a slog.

I enjoyed this and did not find it a slog. There is a (very narrow) chain of logic that’ that leads from the start to the end and avoids too many repetitive calculations. This relies on using each piece of information that we are given at the appropriate time to make the next logical step.

I have now solved it three times and each time I feel I have gotten nearer to the “ideal” path from start to end. A very nice construction with *just* enough information provided to make it solvable.

Thanks for this, Roma.
I have a list of potential squares but I did not spot that the three clues are all on the bottom row of the three grids. D'oh!
I also have an idea now how the value of x can be easily deduced.
Onwards and upwards...

I scanned the puzzle and printed it before starting - big mistake!!
I read the F1 and R2 as having division signs not additions!! Incredibly I almost finished it but with just one parity error! Then started again and found it a lot more of a slog and gave up after completing the first grid! I don't see how anyone could do this without huge amounts of time spent sifting through Excel spreadsheets!

The problem with numericals is that they are hard to pick up and put down. I find I end up going back and rechecking what I have already put in. There is a sequence to this one which keeps options fairly limited. I guess we shouldn't be surprised that it requires a bit of effort. The normal puzzles take me far longer and definitely require some inside knowledge. The numericals are a bit more of a level playing field, I would say. Anyway, pleased to get one completed after two weeks of half filled grids.

This has finally sent me over the edge. Against my better judgement I decided to have one last go at this and I have 3 full grids that satisfy all requirements other than R2. My entry is not divisible by X, and is therefore incorrect, but is only 20 higher than what is the correct R2 based on the sum of all my digits. The sum of my digits in the front grid is 25 less than the top grid and 9 less than the right grid and I cannot see what I have done wrong - unless there is a 2nd solution to all the other clues.

R2 is not divisible by X. It is divisible by (X+Y). Not the same thing at all

I mean divisible by (X+Z), of course

Sorry Dodgepot - my brain is scrambled. I meant F3 not R2.

Ok. Your digit sums for each grid are correct, so you know the correct answer for F3. If yours is 20 higher, then your third digit in F3 is 4. But the third digit in R3 should be 4. You probably have two digits transposed somewhere,