[Guest]

take 3 numbers a, b, c, take the average of each pair, (a, b; b, c; a, c) and the average of all 3.

divide each pair average by the average of all 3, and total. this will be the score. what's the maximum possible score, if a, b, c are distinct integers less than 1,000?

edit: this was a bad puzzle as the answer is always 3.

hmmm, sorry. let me rethink this.

Edited January 7, 2011 by phillip1882

[unreality]

take 3 numbers a, b, c, take the average of each pair, (a, b; b, c; a, c) and the average of all 3.

divide each pair average by the average of all 3, and total. this will be the score. what's the maximum possible score, if a, b, c are distinct integers less than 1,000?

m1 = (a+b)/2

m2 = (b+c)/2

m3 = (a+c)/2

M = (a+b+c)/3

score = m1/M + m2/M + m3/M

= (3/2)(a+b)/(a+b+c) + (3/2)(b+c)/(a+b+c) + (3/2)(a+c)/(a+b+c)

= 3/(2(a+b+c)) * (a + b + b + c + a + c)

= (3/2)/(a+b+c) * (2a + 2b + 2c)

= (2)(3/2)(a+b+c)/(a+b+c) = 3

This would suggest that the 'score' is always 3. Let me check that:

40, 200, 6

m1 = 120

m2 = 103

m3 = 23

M = 82

120/82 + 103/82 + 23/82 = 3, yea..

So unless I'm not understanding the problem description, the score is always 3

edit; yea nvm didn't see your edit sorry

Edited January 7, 2011 by unreality

[Guest]

okay how about this; take three numbers a, b, c. average each pair. you get 3 new numbers. average each pair again. etc. find a formula for the nubmer you apraoch in terms of a, b, and c.

[Guest]

(a +b +c)/3.

it's hard to come up with a good puzzle!