If

E = (F – I + R)
F = (N * O)
H = (F + X)
I = (N + O)
N = 3
0 = 2
R = (N – O)
T = (X + Y)
U = (O + O)
W = (Y + R)

when does

     (T+H+I+R+T+E+E+N)-(T+H+R+E+E)
     ----------------------------- = (T+W+0)  ?
          (N+I+N+E)-(F+O+U+R) 

(All Os ARE the letter Os, not zeros)

Assume that every letter stands for exactly one number, and that the arithmetic operations work the way they're expected to. Then doing some substitutions, we get:
F = 3 * 2 = 6
I = 3 + 2 = 5
R = 3 - 2 = 1
E = 6 - 5 + 1 = 2
U = 2 + 2 = 4
H = 6 + X
T = X + Y
W = 1 + Y

Substituting into just the denominator, we get
(N+I+N+E)-(F+O+U+R) = (3+5+3+2)-(6+2+4+1) = 13 - 13 = 0

Since this leads to a division by zero, I'd be tempted to say no.

Posted by friedlinguini on 2002-12-19 03:15:22