A wheel of fortune has the integers from 1 to 25 placed on it in a random manner. show that regardless of how the numbers are positioned on the wheel, there are three adjacent numbers whose sum is at least 39.

Suppose the 25 integers are arranged a, b, c, ..., y around the wheel, where "a" is adjacent to "y". We will prove by contradiction. For the contradiction, suppose every three adjacent numbers has sum < 39 Then we have this set of 25 inequalities. We add the left and right sides of them all. Notice there each letter appears in 3 inequalities, there is one "a" in the 1st inequality and 2 "a"'s in the last two. Also there are 2 "b"'s in the first two inequalities and 1 "b" in the last inequality. That is, a+b+c < 39 b+c+d < 39 c+d+e < 39 d+e+f < 39 ... v+w+x < 39 w+x+y < 39 x+y+a < 39 y+a+b < 39 Add up all the inequalities. 3a+3b+3c+3d + ... + 3v+3w+3x+3y < 39·25 3 (a+b+c+d + ... + v+w+x+y) < 975 Divide both sides by 3 a+b+c+d + ... + v+w+x+y < 325 However the sum of the integers from 1 through 25, inclusive, is given by the formula: n (n+1) / 2 where n = 25 25 (25+1) / 2 = 325 So the sum (a+b+c+d + ... + v+w+x+y) cannot be both less than 325 and also equal to 325. So the assumption that all 25 of those inequalities hold is false. So there must be at least one of the inequalities which is incorrect. Therefore there are three adjacent numbers whose sum is at least 39.