The mean of six positive integers is 15. The median is 18, and the onl...
Given:
- Mean of six positive integers = 15
- Median = 18
- Mode is less than 18
To find:
The maximum possible value of the largest of the six integers
Solution:
Let's assume the six positive integers as a, b, c, d, e, and f, where a ≤ b ≤ c ≤ d ≤ e ≤ f.
Mean of six positive integers:
Mean is the sum of all the numbers divided by the total count.
Mean = (a + b + c + d + e + f)/6
Given that the mean is 15, we have the equation:
(a + b + c + d + e + f)/6 = 15
a + b + c + d + e + f = 90 ...(Equation 1)
Median:
The median is the middle number when the numbers are arranged in ascending order.
Since the median is 18, we have three possible cases:
Case 1: If c = d = 18
Case 2: If b = c = 18
Case 3: If a = b = 18
Mode:
The mode is the number that appears most frequently in the given set of numbers.
Given that the mode is less than 18, we have two possible cases:
Case 1: a < />
Case 2: b < />
Now, let's consider all the possible cases to find the maximum value of the largest number.
Case 1: If c = d = 18
Given that c = d = 18, the equation (Equation 1) becomes:
a + b + 36 + 36 + e + f = 90
a + b + e + f = 18 ...(Equation 2)
Since the mode is less than 18, we have two sub-cases:
Sub-case 1: a < />
In this case, the maximum value of b + e + f will occur when b = e = f = 17. (b, e, and f should be as close to 18 as possible without exceeding it)
So, a + 17 + 17 + 17 = 18
a + 51 = 18
a = -33 (which is not a positive integer)
Sub-case 2: a = 18
In this case, the maximum value of b + e + f will occur when b = e = f = 17. (b, e, and f should be as close to 18 as possible without exceeding it)
So, 18 + 17 + 17 + 17 = 18
69 = 18 (which is not possible)
Hence, this case is not valid.
Case 2: If b = c = 18
Given that b = c = 18, the equation (Equation 1) becomes:
a + 36 + 36 + d + e + f = 90
a + d + e + f = 18 ...(Equation 3)
Since the mode is less than 18, we have two sub-cases:
Sub-case 1: a < />
In
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