Identify the problem type. It's a mean, median and mode problem, and all the numbers are positive integers. What do you know about all of this?

First they are all positive integers. None are 0, because it's neither positive nor negative.

mean = sum/(number of numbers) which in this case is 15 = sum/6 so sum = 90

median - is the middle of the numbers. When there's an even number of numbers the median is the average of the middle two numbers. In this case that means the middle numbers are either both 18 (x, y, 18, 18, w, z), or they are 17 and 19 (x, y, 17, 19, w, z).

mode - the number repeated the most, and a group of numbers can have multiple modes. The mode is less than 18, which means the middle numbers have to be 17 and 19. If the two middle numbers were 18, then 18 would be a mode because there are only two numbers that can be below 18 (x and y). If those numbers are the same, then both 18 and x,y would be modes. 18 can't be a mode because it isn't less than 18. It wouldn't matter to the average because 18+18 is the same as 17+19.

It does matter to w and z, however. Since the only mode is less than 18, 19 can't repeat. So w has to be at least 20.


To maximize the largest of the numbers, minimize all the other numbers.

x,y,17,19,w,z

x and y can be 1, so the mode is 1 (less than 18). That's minimizing x and y.

1,1,17,19,w,z

20 is the lowest possibility for w. If w were 19, then 19 would be a mode. 19 is not less than 18.

1,1,17,19,20,z

Now find z, which will be the largest possibility since all the other numbers are as small as they can be.

1+1+17+19+20+z (the sum) = 90

58+z=90
z = 90-58=32

1,1,18,18,19,32 could be the feasible solution but in any case max value could be 32

vipulgoyal wrote:1,1,18,18,19,32 could be the feasible solution but in any case max value could be 32

Hey,

The question states that the mode has to be less than 18. If our six numbers are 1,1,18,18,19,32..Your mean comes to less than 15. Furthermore your mode is 1 and 18 which is not feasible. In the case you have mentioned, the maximum value comes to 33, not 32, if we are to get a mean of 15.

1,1,18,18,19,32 - Did you add those numbers and divide the sum by 6? The mean is 14.833... There are two modes, 1 and 18.

I may be wrong, but can we have these series.

1,1,17,19,1,51 OR 1,1,18,18,1,51

1,1,17,19,1,51 OR 1,1,18,18,1,51

The median of each set is not 18. Order the list. The median of an even number of numbers is the mean of the middle two numbers.

1, 1, 1, 17, 19, 51 - The median is 9. (1+17)/2 = 18/2 = 9
1,1,1,18,18,51 - the median is 9.5. (18+1)/2 = 9.5

fcabanski wrote:Identify the problem type. It's a mean, median and mode problem, and all the numbers are positive integers. What do you know about all of this?

First they are all positive integers. None are 0, because it's neither positive nor negative.

mean = sum/(number of numbers) which in this case is 15 = sum/6 so sum = 90

median - is the middle of the numbers. When there's an even number of numbers the median is the average of the middle two numbers. In this case that means the middle numbers are either both 18 (x, y, 18, 18, w, z), or they are 17 and 19 (x, y, 17, 19, w, z).

mode - the number repeated the most, and a group of numbers can have multiple modes. The mode is less than 18, which means the middle numbers have to be 17 and 19. If the two middle numbers were 18, then 18 would be a mode because there are only two numbers that can be below 18 (x and y). If those numbers are the same, then both 18 and x,y would be modes. 18 can't be a mode because it isn't less than 18. It wouldn't matter to the average because 18+18 is the same as 17+19.

It does matter to w and z, however. Since the only mode is less than 18, 19 can't repeat. So w has to be at least 20.


To maximize the largest of the numbers, minimize all the other numbers.

x,y,17,19,w,z

x and y can be 1, so the mode is 1 (less than 18). That's minimizing x and y.

1,1,17,19,w,z

20 is the lowest possibility for w. If w were 19, then 19 would be a mode. 19 is not less than 18.

1,1,17,19,20,z

Now find z, which will be the largest possibility since all the other numbers are as small as they can be.

1+1+17+19+20+z (the sum) = 90

58+z=90
z = 90-58=32

If median is 18, middle numbers doesnt necessarily be 18,18 or 17,19. It could be any two numbers whose sum is 36.

I am hoping someone to suggest a fast method to approach such problems.

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

"If median is 18, middle numbers doesnt necessarily be 18,18 or 17,19. It could be any two numbers whose sum is 36. "

If median were the only restriction, that would be true. But since the mode is less than 18, and the other numbers have to be as small as possible so the missing number is as large as possible, the middle integers have to be 17 and 19.

It takes about 20 seconds to solve this problem. It takes longer to explain how to solve it.