AMC 10 · 2022 · #6

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, one of them unknown. The mean of the six equals one of the values already appearing in the set. Find every positive integer the unknown can be, and add those values together.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

Add the known numbers

Find the sum of the five known numbers.

1+7+5+2+5 = 20 → mean = (20+X)/6

2STEP 2

Split into cases

Case on which value the mean equals.

cases: mean ∈ {1, 2, 5, 7, X}

3STEP 3

The small values fail

Small values give a negative answer.

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

4STEP 4

The first success

One value gives a positive integer.

X = 6 · 5 - 20 = 10 ✓

5STEP 5

The second success

Another value also works.

X = 6 · 7 - 20 = 22 ✓

6STEP 6

The mean equals the unknown

It can also equal the unknown itself.

20 + X = 6X → X = 4 ✓

7STEP 7

Add them up

Adding them gives 36.

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 12 problem only needs Grade 6 mean and one-step equations you already know — write the average as (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.