OT: Interesting math problem...

[Will]

A more rigorous "proof", took me about five minutes before I went to class, posting it now after class:

Code:

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 TRIED

+DRIVE

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 RIVET



Assuming we disregard the trivial solution (all zeros), and 

force each letter to be distinct (ie for any x and y in 

{T,R,I,E,D,V}, the values x != y), and noticing that the 

carry can be at most 1 (9 + 9 + 1 = 19):



either E + V = E (V = 0)

    or E + V + 1 = E (V = 9)

disregard V = 0 as uninteresting, so V = 9

    and D + E > 10 and T < D, T < E

either I + I = V = 9

    or I + I + 1 = V = 9

I + I != 9, so I + I + 1 = V is true

so I = {4,9}, but V = 9, so I = 4

so R + R = I = 4 (no carry from previous digit)

then R = {2,7}

V = 9, I = 4, R = {2,7}

either T + D = R = 2

    or T + D + 1 = R = 7

assume there is no final carry

so if T + D = 2, then T = D = 1

then T + D + 1 = 7 = R

then T < (7 - D)

V = 9, I = 4, R = 7, T < (7 - D), T < E

try T = 1:

V = 9, I = 4, R = 7, T = 1, D = 5, E = 6

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