## 2014 AMC 10B Problem 18 / AMC 12B Problem 11

2014 AMC 10B 18 / AMC 12B 11. A list of 11 positive integers has a mean of 10, a median of 9, and a unique mode of 8. What is the largest possible value of an integer in the list?

(A) 24   (B) 30   (C) 31   (D) 33   (E) 35

Let’s start by naming the positive integers with some notation so that we can manipulate them. Let these 11 positive integers be $0 < x_1 \leq x_2 \leq \dots \leq x_{11}.$ The question then becomes what the largest value of $x_{11}$ can be under the given conditions, meaning that we should try to make the rest of the variables as small as possible.

Since the mean of the $x_i$‘s is 10, we have the equation

$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} + x_{11} = 110.$

The number right in the middle is $x_6$, so taking the median condition into account, the equation above becomes

$x_1 + x_2 + x_3 + x_4 + x_5 + 9 + x_7 + x_8 + x_9 + x_{10} + x_{11} = 110.$

The last condition (“unique mode is 8”) is the trickiest to work with. The first thing to note is that since the mode must be unique and there is already one 9 among the integers, there must be at least two ‘8’s in the list:

$x_1 + x_2 + x_3 + 8 + 8 + 9 + x_7 + x_8 + x_9 + x_{10} + x_{11} = 110.$

(“Wait!” you might say. “How do you know that $x_5$ must be 8? Couldn’t $x_5 = 9$? That would still fulfill the increasing nature of the integers and the median condition.” True, but in that case, the numbers in the sequence would generally be larger and $x_{11}$ would not be maximised. Challenge: Prove this in a watertight way.)

At this point there may be other ways to proceed, but let me show you how I did it– by looking at all possibilities for the number of 8s in the $x_i$‘s. Once the number of 8s are fixed, we then make $x_1, \dots, x_{10}$ as small as possible (but still fulfilling the conditions).

Case 1: There are exactly 2 8s.
To maximise $x_{11}$, we have

\begin{aligned} 1 + 2 + 3 + 8 + 8 + 9 + 10 + 11 + 12 + 13 + x_{11} &= 110, \\ x_{11} &= 33. \end{aligned}

Case 2: There are exactly 3 8s.
To maximise $x_{11}$, we have

\begin{aligned} 1 + 1 + 8 + 8 + 8 + 9 + 9 + 10 + 10 + 11 + x_{11} &= 110, \\ x_{11} &= 35. \end{aligned}

Since this is a multiple choice question and 35 is the largest possible answer given in the options, we can stop here and conclude that the answer is (E).

For completeness’ sake, below are the remaining cases that we should have gone through if the question wasn’t multiple choice:

Case 3: There are exactly 4 8s.
To maximise $x_{11}$, we have

\begin{aligned} 1 + 8 + 8 + 8 + 8 + 9 + 9 + 9 + 10 + 10 + x_{11} &= 110, \\ x_{11} &= 30. \end{aligned}

Case 4: There are exactly 5 8s.
To maximise $x_{11}$, we have

\begin{aligned} 8 + 8 + 8 + 8 + 8 + 9 + 9 + 9 + 9 + 10 + x_{11} &= 110, \\ x_{11} &= 24. \end{aligned}