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2022 AMC 10A Problem 8

2022 AMC 10A Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Statistics & Data · difficulty 2 of 5 · about 3 min · position P1-10

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 $X$ ?

A)
$10$
B)
$26$
C)
$32$
D)
$36$
E)
$40$

Answer

Key insight

The mean (20 + X)/6 must equal 1, 2, 5, 7, or X itself; only 5, 7, and X yield a positive integer X, namely 10, 22, and 4.

Solution

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 $\boxed{\textbf{(D)}\ 36}$ .

Why this works

"The mean equals a value in the set" is a finite list of equations, one per candidate value. The unknown itself is one of those candidates, which is the case most easily overlooked. Solve every case, then discard solutions that violate the stated constraints (here, positivity).

The trap

Forgetting the case where the mean equals X itself (X = 4), which gives the sum 32, choice (C).

Common mistakes

Forgetting the case where the mean equals X itself (X = 4), which gives the sum 32, choice (C).
Including the negative solutions $-14$ and $-8$ , or treating the two $5$ s as separate cases and counting $X = 10$ twice.

Techniques

Split into exhaustive cases and handle each