Will, that's a good thourough proof there, but I'm curious why you "discount V=0 as uninteresting". Maybe that's a math thing I don't get, but it seems V=0 is just as valid a posibility as V=9 until it's proven or disproven. In fact I did disprove V=0 in this case, in a very similer fasion to how you proved V=9.
If V=0 then
D + E < 10, since no number for E + 0 + 1 = E or 10 + E
Since D + E < 10, there is no carry over to the second or third columns.
Therefore I = 5, since I must be either 0 or 5 and we have already determined V = 0.
I + I gives us a carry to the next position so R + R + 1 = 5. For this to be correct R must be either 2 or 7.
(Somehow I got this point while disproving V = 0, but forgot it while attempting to prove V = 9 ) If R is 2 then there is no carry and T + D is 2. For this to be correct either T and D must both be 1 or T and D are 0 and 2. Both violate our assumptions so R cannot be 2, therefore R = 7.
Therefore 1 + T + D = 7, T + D = 6. T and D cannot both be 3, and neither T or D can be 0(+6) or 5(+1), so T & D must be 2 & 4. To determine which is which we can try and substitute that into D + E = T (Which as we have already calculated is < 10)
If D is 2, E would have to also be 2 for T to be 4, so D cannot be 2.
If D is 4, E would have to be -2 for T to be 2, so D cannot be 4.
Since D cannot be 2 or 4 then it follows that V cannot be 0
Heh, reading this out loud I realized I sound like Vizzini from the Princess Bride.
Re: OT: Interesting math problem...
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