2022 AMC 10A Problem 8

P1-10~3 minPrint

Statistics & Data Split into exhaustive cases and handle each

A data set consists of $6$ (not distinct) positive integers: $1$ , $7$ , $5$ , $2$ , $5$ , and $X$ . The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all possible values of $X$ ?

Show solution without answering

Key insight

The mean is (20 + X)/6. Set it equal to each number in the list, including X itself, and keep the answers where X is a positive whole number.

Solution

The five known numbers add to $1 + 7 + 5 + 2 + 5 = 20$ . So the mean of all six numbers is $\dfrac{20 + X}{6}$ .
The mean has to be one of the numbers in the list: $1$ , $2$ , $5$ , $7$ , or $X$ itself. Try each one. If the mean equals some value, then $20 + X = 6 \times \text{value}$ .

Mean equals	Equation	$X$	Allowed?
$1$	$20 + X = 6$	$-14$	no, not positive
$2$	$20 + X = 12$	$-8$	no, not positive
$5$	$20 + X = 30$	$10$	yes
$7$	$20 + X = 42$	$22$	yes
$X$	$20 + X = 6X$	$4$	yes

The last row needs a word: $20 + X = 6X$ gives $20 = 5X$ , so $X = 4$ . Check: the list $1, 7, 5, 2, 5, 4$ adds to $24$ , its mean is $4$ , and $4$ is in the list.
The possible values are $10$ , $22$ and $4$ , and their sum is $36$ .

The answer is $\boxed{\textbf{(D)}\ 36}$ .

Why this works

"The mean equals a number in the list" is really a short list of equations, one for each number. The unknown $X$ is in the list too, so it gets its own equation, and that is the one most people miss. After solving, throw out any answer that breaks the rules (here, $X$ must be positive).

Concise solution

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

The five known numbers sum to $20$ , so the mean is $\dfrac{20 + X}{6}$ . It must equal one of the data values $1, 2, 5, 7$ , or $X$ .

Mean $= 1$ : $20 + X = 6$ , $X = -14$ . Not positive.
Mean $= 2$ : $20 + X = 12$ , $X = -8$ . Not positive.
Mean $= 5$ : $20 + X = 30$ , $X = 10$ . Valid.
Mean $= 7$ : $20 + X = 42$ , $X = 22$ . Valid.
Mean $= X$ : $20 + X = 6X$ , $X = 4$ . Valid (the mean $4$ is in the set because $X = 4$ is).

The possible values are $10, 22, 4$ , with sum $36$ .

The answer is $\boxed{\textbf{(D)}\ 36}$ .

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)