AMC 8 · 2015 · #13

Grade 7 arithmeticcounting

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

mean-median-mode-rangesequences-arithmeticsystematic-enumeration systematic-enumerationidentify-subproblems ↑ Prerequisites: mean-median-mode-rangemulti-digit-arithmetic

📏 Short solution 💡 3 insights

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Problem

From the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, remove a two-element subset {a, b}. How many such subsets leave the remaining 9 numbers with an average (mean) of 6?

Pick an answer.

(A)

$text{ 1}$

(B)

$text{ 2}$

(C)

$text{ 3}$

(D)

$text{ 5}$

(E)

$text{ 6}$

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

We don't need to test every pair. Tool #7 (Identify Subproblems) splits the question into three small steps: (1) sum of S, (2) required sum of the kept 9 numbers, (3) sum of the removed pair. Tool #13 (Work Backwards) is the engine for step 3 — knowing the target mean tells us the sum that must remain, which forces the sum of the removed pair. Then Tool #6 (Make an Organized List) finishes the job: list every pair of distinct elements in S that adds to that forced sum.

1STEP 1

Add up the whole set: the eleven consecutive integers sum to 66.

1 + 2 + … + 11 =

\frac{11 \cdot 12}{2}

= 66

2STEP 2

The mean is the sum over the count, so the nine kept numbers must total 54.

Sum_kept = 6 × 9 = 54

3STEP 3

Subtract to see the removed pair must carry the difference, a sum of 12.

Sum_removed = 66 - 54 = 12

4STEP 4

List every pair of distinct numbers in the set that adds to 12, walking the smaller value up from 1.

{1, 11}, {2, 10}, {3, 9}, {4, 8}, {5, 7}

5STEP 5

Reject {6, 6} because a subset needs distinct elements, leaving five valid pairs.

Valid pairs: {1,11}, {2,10}, {3,9}, {4,8}, {5,7} — 5 pairs → (D)

Answer

text{ 5}

S is symmetric around its middle value 6, and the kept mean (6) is exactly that middle value. So we should be removing pairs whose own mean is also 6 — i.e. pairs that sum to 12. The pairs {1,11}, {2,10}, {3,9}, {4,8}, {5,7} are exactly the 5 symmetric pairs around 6, with the "pair" {6, 6} correctly excluded because subsets need distinct elements. Answer (D) = 5 matches.

💡Key takeaway

This AMC 8 problem really comes down to one Grade 6 idea — average = sum ÷ count — plus careful pair-counting you already know from Grade 7.

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