Replace each letter by a positive rational number such that the following is true. Do it for rationals that can be written in the same denominator that is as small as possible.

O + N + E          =  1
T + W + O          =  2
T + H + R + E + E  =  3
F + O + U + R      =  4
F + I + V + E      =  5
S + I + X          =  6

(In reply to to goFish and dopey - about 0 by pcbouhid)

For distinct positive rationals, finding

O + N + E = 1 is equivalent to finding a/k + b/k + c/k =1

or a + b + c = k for distinct positive integers. It should be clear that 1 + 2 + 3 = 6 is the smallest possible answer.

ps.  I too did not think originally that the solution required distinct solutions.  In fact, I thought the omission of some indication to that effect in the problem, implied distinct solutions were not required.

 

Posted by goFish on 2006-01-02 11:46:58