vipulgoyal wrote:The mean of six Positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the largest of the six integers is
Ans 32[spoiler][/spoiler]
Let's turn this into a system of equations, and say that our set is a, b, c, d, e, and f. Without loss of generality, we'll say that c and d are the two "middle numbers". (In other words, a ≥ b ≥ c ≥ d ≥ e ≥ f.)
Since the mean of the six integers is 15, we have
(a + b + c + d + e + f)/6 = 15, or
(a + b + c + d + e + f) = 90
Since the median is 18, we have
(c + d)/2 = 18, or
c + d = 36
The tricky part is the mode. Since we have a mode, we must have at least two numbers that are equal. Since the mode is less than 18, those numbers can only be two or three of d, e, and f. (c can't be one of them, since c ≥ 18. d might also = 18, but only if c = d.) In equations, we have either d = e = f, or d = e ≠f or d ≠e = f.
Since we're looking for the maximum value of a, we want to minimize the value of everything else. Since c + d = 36, we'll make c = 19 and d = 17. (They can't be equal, or we'll have a mode of 18!) We'll go even further and say that b = 20, minimizing b. (b can't equal 19, or we'd have a mode of 19.) Since e and f must be positive, we'll make e = f = 1, the smallest positive integer and hence the minimum value of e and f.
a + b + c + d + e + f = 90, so
a + 20 + 19 + 17 + 1 + 1 = 90, so
a = 32
