2022 AMC 10B Problem 10

P1-10~5 minPrint

Statistics & Data Number Properties Bound the quantity above/below or estimate to pin it down Consider the largest/smallest element or boundary case

Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?

Show solution without answering

Key insight

The mode is above the median, so it fills the top two spots. The two smallest numbers must differ, which forces the mean to be at least 7.

Solution

Give the mean a name and write everything else in terms of it. Say the mean is $m$ . Then the median is $m + 2$ and the mode is $m + 4$ .

1. The mean is a whole number. With five numbers, the median is the middle one, so it is a whole number. The mean is $2$ less, so it is a whole number too.
2. Where the mode sits. List the numbers from smallest to largest: $a, b, c, d, e$ . The median is $c = m + 2$ . The mode is bigger than $c$ , so it can only be $d$ or $e$ . A mode appears at least twice, so $d = e = m + 4$ .
3. No other repeats. The mode is the only one, so no other value may appear twice. So $a$ , $b$ and $c$ are all different, which means $a < b < c$ .
4. Use the total. Five numbers with mean $m$ add up to $5m$ :
$a + b + (m + 2) + (m + 4) + (m + 4) = 5m,$
which simplifies to $a + b = 2m - 10$ .
5. Smallest $m$ (cannot go lower). The smallest two different positive whole numbers are $1$ and $2$ , so $a + b \ge 3$ . Then $2m - 10 \ge 3$ , so $m \ge 6.5$ . Since $m$ is a whole number, $m \ge 7$ .
6. $m = 7$ works. Then $a + b = 4$ with $a < b$ , so $a = 1$ , $b = 3$ . The list is $1, 3, 9, 11, 11$ . Check: the sum is $35$ and the mean is $7$ ; the median is $9$ ; the only repeated value is $11$ . Everything fits.
7. The mode is $m + 4 = 11$ .

The answer is $\boxed{\textbf{(D)}\ 11}$ .

Why this works

Turn each word into a fact about the sorted list: the mean fixes the total, the median is the middle entry, and a single mode above the median has to take the top two spots. After that, making the answer small is just a matter of making the two smallest entries as small as the rules allow. Do not forget that the median of an odd number of whole numbers is itself a whole number.

Alternative approach

Test the answer choices from the smallest. Mode $5$ means mean $1$ , but five positive whole numbers with mean $1$ are all $1$ , so the mode would be $1$ . Mode $7$ means mean $3$ and total $15$ ; the numbers $5, 7, 7$ already add to $19$ , too much. Mode $9$ means mean $5$ and total $25$ ; the numbers $7, 9, 9$ add to $25$ , leaving $0$ for $a + b$ , impossible. Mode $11$ works with $1, 3, 9, 11, 11$ .

Concise solution

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

Let the mean be $m$ , so the median is $m + 2$ and the mode is $m + 4$ . The median of five integers is the middle integer, so $m + 2$ and hence $m$ are integers.

Sort the numbers $a \le b \le c \le d \le e$ , with $c = m + 2$ . The mode $m + 4$ is larger than the median, so it can only occupy positions $d$ and $e$ ; being a mode it appears at least twice, so $d = e = m + 4$ .

Since the mode is unique, no other value may appear twice. Thus $a < b < c$ (if $b = c$ , the median value would also appear twice).

The sum is $5m$ :
$a + b + (m+2) + 2(m+4) = 5m \quad\Longrightarrow\quad a + b = 2m - 10.$
With $1 \le a < b$ , the smallest possible $a + b$ is $1 + 2 = 3$ , so $2m - 10 \ge 3$ , i.e. $m \ge 7$ (as $m$ is an integer).

For $m = 7$ : $a + b = 4$ , so $(a, b) = (1, 3)$ and the list is $1, 3, 9, 11, 11$ . Check: mean $35/5 = 7$ , median $9$ , unique mode $11$ . The mode is $11$ .

The answer is $\boxed{\textbf{(D)}\ 11}$ .

Review in the AoPS books

Prealgebra, Chapter 13: Data and Statistics
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 20: Special Functions (§20.3)