new puzzle challenge

[Gilles]

(04-01-2015 12:25 PM)Claudio L. Wrote:  Wait... not so fast.
Adding all 15 equations, you get each letter 3 times exactly, so:
3*(A+B+....+R+S)=38*15
3*(A+B+....+R+S)=38*3*5
(A+B+....+R+S)=38*5=190
So that extra equation is not linearly independent.
We still have 7 variables. Back to the drawing board.

Yes. You're right...
A+B+C=38
D+E+F=38
etc..
(A+B+C)+(D+E+F)+.....= 5x38

I wrote a rpl program that explores 160.392.960 cases 'only' and use the solve command. But too slow to be usable even with emulator