AMC 10 · 2022 · #8

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

mean-median-mode-rangecaseworkdivisibility-rules caseworksystematic-enumeration ↑ Prerequisites: mean-median-mode-range

📏 Long solution 💡 3 insights

Problem

A data set has six positive integers {1, 7, 5, 2, 5, X}. The arithmetic mean of the six numbers equals one of the values that already appears in the data set. Find every positive integer X that makes this happen, and add those X-values together.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The phrase "equals a value in the data set" gives a small finite list of cases: the mean can be 1, 2, 5, 7, or X. Tool #2 (Systematic List) walks through every case in order so nothing is missed. Within each case Tool #11 (Work Backwards) finds X — we know the mean, so we run the average formula backwards to recover the missing piece. Tool #6 (Guess and Check) is the verification: plug the candidate X back into the full data set and confirm the recomputed mean really does land in the set.

1STEP 1

The five fixed numbers total 1+7+5+2+5=20, so the mean of all six is $\frac{20+X}{6}$ .

1+7+5+2+5 = 20 → mean =

\frac{20+X}{6}

2STEP 2

The distinct in-set values are {1, 2, 5, 7} plus the wildcard X, giving five cases: mean = 1, 2, 5, 7, X.

cases: mean ∈ {1, 2, 5, 7, X}

3STEP 3

mean=1 gives X=-14 and mean=2 gives X=-8 — both negative, so rejected; a positive X can never pull the average down to 1 or 2.

X = 6· 1 - 20 = -14, X = 6· 2 - 20 = -8

4STEP 4

mean=5 gives X=10 — a positive integer; {1,7,5,2,5,10} has mean 5, and 5 is in the set ✓.

X = 6· 5 - 20 = 10 ✓

5STEP 5

mean=7 gives X=22 — a positive integer; {1,7,5,2,5,22} has mean 7, and 7 is in the set ✓.

X = 6· 7 - 20 = 22 ✓

6STEP 6

mean=X: 20+X=6X → 5X=20 → X=4, a positive integer that is in the set by construction ✓.

20 + X = 6X → X = 4 ✓

7STEP 7

The survivors are X ∈ {4, 10, 22}, and their sum 4 + 10 + 22 = 36 is choice (D).

4 + 10 + 22 = 36 → (D)

Answer

Each surviving X has been independently verified by recomputing the mean of the full six-element set and confirming the value lives in {1,2,5,7,X}. The cases mean=1 and mean=2 were correctly rejected because the five fixed numbers already total 20, forcing X negative. The sum 36 matches choice (D); the nearby choices 32 and 40 would correspond to forgetting one of {4, 22} or including a spurious case.

💡Key takeaway

This AMC 10 problem only needs Grade 6 mean and one-step equations you already know — write the average as $\frac{20+X}{6}$ , try each in-set value for it, keep the cases that give a positive integer, and add the survivors 4+10+22=36.

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