2014 AMC 10B Problem 18

P16-20~6 minPrint

Statistics & Data Split into exhaustive cases and handle each Consider the largest/smallest element or boundary case

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?

Show solution without answering

Key insight

To make the biggest number large, make the other ten as small as possible. Using only three 8s lets two 1s into the low spots.

Solution

1. Total. Eleven numbers with mean $10$ add to $11 \times 10 = 110$ .
2. The goal. The largest number is $110$ minus the other ten. So make the other ten add to as little as possible.
3. The median. In order from smallest to largest, the $6$ th number is $9$ . So five numbers sit at or below $9$ and five sit at or above $9$ .
4. The mode. $8$ must appear more often than any other value. Every $8$ is below the median, so $8$ appears at most $5$ times. Say it appears $k$ times; then every other value appears at most $k - 1$ times.
5. Why not use lots of 8s? Two things pull in opposite directions:
- more $8$ s let us repeat cheap values like $9$ and $10$ above the median;
- fewer $8$ s leave low spots free for very small numbers like $1$ .
So try each $k$ and find the smallest possible total of the ten numbers.

copies of 8	cheapest ten numbers	their sum	largest number
$5$	$8,8,8,8,8,\ 9,\ 9,9,9,10$	$86$	$24$
$4$	$1,8,8,8,8,\ 9,\ 9,9,10,10$	$80$	$30$
$3$	$1,1,8,8,8,\ 9,\ 9,10,10,11$	$75$	$35$
$2$	$1,2,3,8,8,\ 9,\ 10,11,12,13$	$77$	$33$

How each row is built: fill the five low spots with the $8$ s and the smallest numbers allowed (each non- $8$ value used at most $k - 1$ times); then the median $9$ ; then the four upper spots with the smallest numbers at least $9$ , again each value used at most $k - 1$ times in total.
(With only one $8$ , no value could repeat, but then $8$ would not be the unique mode, so $k \ge 2$ .)
6. The smallest total is $75$ , with three $8$ s. The list $1, 1, 8, 8, 8, 9, 9, 10, 10, 11, 35$ adds to $110$ , has median $9$ , and $8$ is the only value that appears three times.

The answer is $\boxed{\textbf{(E)}\ 35}$ .

Why this works

With a fixed mean, the total is fixed, so "make one number as big as possible" means "make the rest as small as possible." The mode rule works both ways: more copies of the mode let other values repeat, but they also take up cheap low spots. With only a few choices for how many $8$ s to use, trying every case is quick and certain.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The total is $11 \cdot 10 = 110$ . Sort the list $a_1 \le \cdots \le a_{11}$ ; the median gives $a_6 = 9$ . To make $a_{11}$ as large as possible, make $a_1 + \cdots + a_{10}$ as small as possible.

Say $8$ appears $k$ times. All the 8s sit below the median, so $k \le 5$ , and every other value appears at most $k - 1$ times. The value $9$ already appears once (as $a_6$ ), and each entry above the median is at least $9$ .

$k = 5$ : $a_1..a_5 = 8$ ; then $9$ at most four times, so $a_7, a_8, a_9 = 9$ , $a_{10} = 10$ . Sum of ten: $40 + 36 + 10 = 86$ , so $a_{11} = 24$ .
$k = 4$ : bottom $1, 8, 8, 8, 8$ ; $a_7, a_8 = 9$ , $a_9, a_{10} = 10, 10$ . Sum $33 + 27 + 20 = 80$ , so $a_{11} = 30$ .
$k = 3$ : bottom $1, 1, 8, 8, 8$ ; $a_7 = 9$ , $a_8, a_9 = 10, 10$ , $a_{10} = 11$ . Sum $26 + 18 + 20 + 11 = 75$ , so $a_{11} = 35$ .
$k = 2$ : bottom $1, 2, 3, 8, 8$ ; $a_7..a_{10} = 10, 11, 12, 13$ . Sum $22 + 9 + 46 = 77$ , so $a_{11} = 33$ .

The best is $k = 3$ : the list $1, 1, 8, 8, 8, 9, 9, 10, 10, 11, 35$ has sum $110$ , median $9$ , and unique mode $8$ .

The answer is $\boxed{\textbf{(E)}\ 35}$ .

Review in the AoPS books

Prealgebra, Chapter 13: Data and Statistics
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Prealgebra, Chapter 14: Counting (§14.3)