poptasticboy wrote: ↑09 January 2024, 15:30
Hi! So a few initial deductions:
E has to be "Sum is a multiple of 3" otherwise B would be redundant.
*
EDIT: this is not actually true! In trying to recreate my reasoning quickly I actually forgot about the case where Sum=10!! For now, we'll *assume* the Sum is a multiple of 3, then I will add a section at the end dealing with why the Sum can't be a multiple of 5!*
F has to be saying that a certain colour is >1.
This is because, if it were saying that a colour is =1, then D would have to be referring to a different colour (otherwise it would be redundant). Then C must be referring to the final unused colour, (otherwise it would be redundant). For a multiple of 3, this only leaves 111 and the other Verifiers would all be redundant.
My two questions told me that:
A) Bl=Y
B) Sum is Odd.
For an Odd Sum multiple of 3 with Bl=Y, the only possible solutions are 111, 225, 441, 333, 555.
We still have C, D and F to do work for us.
F rules out 111. C rules out 555, but we'll keep them in mind for consideration of redundancy.
If C refers to P, we have 111, 441, 333 left.
D then must refer to Bl or Y to not be redundant, ruling out 441.
F rules out 111, leaving 333.
On the other hand, if C refers to Bl or Y, we have 111, 225, 333 left.
D must then refer to P to not be redundant, ruling out 225.
F rules out 111, leaving 333.
Either way, this leaves just 333 as the solution.
*
EDIT: Now for the "Sum is a multiple of 5" case!*
So if the Sum is a multiple of 5, and from my two tests Bl=Y and Sum is Odd, then the solution can only be 113, 221 or 555, and again we have C, D and F to work with.
C just rules out 555, no matter which colour it refers to. D then chooses a unique solution between 113 and 221, leaving F redundant.
Therefore the Sum can not be a multiple of 5, and the previous solution 333 stands!
Sorry for the initial error!
.