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rdl - perhaps share the sum of the digits of your final sum, which hopefully wont give much away but should help indicate if you have made mistake.

...yes, yes it is. Well I feel stupid now.... I checked to see if the two repeated numbers were triangular, but not the sum.

I'm still confused by the preamble but surely s is the sum of A+B and A+C and B+C and A+B+C = 3(A+B+C). Is that not right?

Cruncher - S is just A + B + C where A, B and C form the thematic triple

Thank you. I parsed the salient comma incorrectly.

Thank you very much, Oyler. I thought that this was a good puzzle. The preamble took a bit of deciphering as others have said, but once I got my head around that it was relatively plain sailing.

I'm not a mathematician although I do have a science degree and a background in geophysics. It was a tough one, and I too struggled with homotopic not appearing in any of my dictionaries. Got there in the end though; I thought it a rather good puzzle with its added twist.

No still lost, even before we get into triangles. Take Clue VII. n m and K are 3 odd numbers each with 2 digits. Largest values are 99,97 and 95. Hence the p q r values are at most 49 48 and 47. So biggest value for s is 144. Hence 16s +3 cannot be bigger than 2307 - which does not have the 5 digits that 2[k]63 requires??

Wezza - the p q and r are the roots of the triangular numbers so in your example you would add
together the 49th, 48th and 47th triangular number to give you s

Sorry that not quite right as they don’t sum to s.

The three integers sum to s and any two of them sum to p, q and r