AMC 8 · 2015 · #13
Grade 7 arithmeticcountingProblem
How many subsets of two elements can be removed from the set so that the mean (average) of the remaining numbers is 6?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: 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 $6$?
Givens: Original set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ ($11$ consecutive integers); Exactly $2$ distinct elements are removed; The mean of the remaining $9$ numbers must equal $6$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $5$, (E) $6$
Unknowns: The number of valid two-element subsets that can be removed
Understand
Restated: 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 $6$?
Givens: Original set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ ($11$ consecutive integers); Exactly $2$ distinct elements are removed; The mean of the remaining $9$ numbers must equal $6$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $5$, (E) $6$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #13 Work Backwards, #6 Make an Organized List
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.
Execute — Answer: D
4.OA.B.4 Step 1 - Add up the original set.
- For $11$ consecutive integers $1, 2, \dots, 11$, use the arithmetic series formula $\tfrac{n(n+1)}{2}$ with $n=11$.
💡 Pairing $1+11, 2+10, \dots, 5+7$ gives five $12$'s plus the middle $6$, which is $5 \cdot 12 + 6 = 66$ — a Grade 4 pattern in arithmetic.
6.SP.B.5 Step 2 - Find the required sum of the $9$ numbers that stay.
- Mean equals total divided by count, so total equals mean times count.
💡 Working backwards from "mean $= 6$ over $9$ numbers" pins down the total — this is the Grade 6 statistics definition of mean.
4.OA.A.3 Step 3 - Subtract to get the required sum of the removed pair.
- Whatever leaves the set must take exactly the difference.
💡 Total = kept $+$ removed, so removed $= $ total $-$ kept. This is a one-step Grade 4 word-problem subtraction.
5.OA.B.3 Step 4 List every pair $\{a, b\}$ of distinct elements of $S$ with $a + b = 12$, walking $a$ upward from $1$.
💡 Walking $a = 1, 2, 3, \dots$ and reading off $b = 12 - a$ is the Grade 5 "generate two related patterns" move.
7.SP.C.8 Step 5 - Reject any pair that breaks the "distinct elements" rule, then count the survivors.
- $6 + 6 = 12$ would need the number $6$ twice, but a subset cannot repeat an element.
- For $a \geq 7$, $b = 12 - a \leq 5 < a$, so we'd just repeat earlier pairs.
💡 Counting unordered pairs of distinct outcomes that meet a condition is the Grade 7 "compound events / sample-space" idea.
4.OA.B.4 Add up the original set. For $11$ consecutive integers $1, 2, \dots, 11$, use th 6.SP.B.5 Find the required sum of the $9$ numbers that stay. Mean equals total divided by 4.OA.A.3 Subtract to get the required sum of the removed pair. Whatever leaves the set mu 5.OA.B.3 List every pair $\{a, b\}$ of distinct elements of $S$ with $a + b = 12$, walkin 7.SP.C.8 Reject any pair that breaks the "distinct elements" rule, then count the survivo Review
Reasonableness: $S$ 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.
Alternative: Tool #11 (Look for Symmetry): the set $\{1,\dots,11\}$ is symmetric about $6$, so its mean is $6$. Removing any pair that is also symmetric about $6$ (i.e. of the form $\{6-k,\ 6+k\}$ for $k = 1, 2, 3, 4, 5$) preserves the mean. That gives $k \in \{1,2,3,4,5\}$, so $5$ pairs — choice (D). $k = 0$ is rejected because it would mean removing $\{6,6\}$, which is not a valid subset.
CCSS standards used (min grade 7)
4.OA.A.3Solve multistep word problems with whole numbers using the four operations (Subtracting $66 - 54 = 12$ to get the required sum of the removed pair from the total and the kept sum.)4.OA.B.4Find factor pairs and recognize arithmetic patterns (Spotting the arithmetic pattern $1 + 2 + \cdots + 11 = \tfrac{11 \cdot 12}{2} = 66$ via pairing.)5.OA.B.3Generate two numerical patterns using given rules (Walking $a = 1, 2, 3, \dots$ and reading off $b = 12 - a$ to enumerate candidate pairs in order.)6.SP.B.5Summarize numerical data sets using measures of center (mean) (Using $\text{mean} = \tfrac{\text{sum}}{\text{count}}$ in reverse: $\text{sum} = 6 \times 9 = 54$ for the kept numbers.)7.SP.C.8Find probabilities of compound events by listing the sample space (Counting unordered pairs $\{a, b\}$ of distinct elements with $a + b = 12$ by exhausting the sample space.)
⭐ This AMC 8 problem really comes down to one Grade 6 idea — average $=$ sum $\div$ count — plus careful pair-counting you already know from Grade 7.
⭐ This AMC 8 problem really comes down to one Grade 6 idea — average $=$ sum $\div$ count — plus careful pair-counting you already know from Grade 7.