AMC 8 · 2015 · #13
Easy mode Grade 7Problem
Start with the set of numbers {1,2,3,4,5,6,7,8,9,10,11}.
Now pick any two of these numbers and remove them. That leaves 9 numbers behind. Take the average of those 9 remaining numbers.
We want the average to come out to exactly 6.
How many different pairs of numbers can we remove so that the average of what is left is 6?
Pick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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.
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