AMC 10 · 2002 · #21
Grade 6 arithmeticThe mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
Pick an answer.
AMC 10 2002 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: A collection of eight integers has its mean, median, unique mode, and range all equal to $8$. Find the largest value that any single integer in the collection could be.
Givens: There are exactly eight integers in the collection; The mean of the eight integers is $8$; The median of the eight integers is $8$; The unique mode is $8$ (the value $8$ appears more often than any other value); The range (largest minus smallest) is $8$; Answer choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Unknowns: The greatest possible value of the largest integer in the collection
Understand
Restated: A collection of eight integers has its mean, median, unique mode, and range all equal to $8$. Find the largest value that any single integer in the collection could be.
Givens: There are exactly eight integers in the collection; The mean of the eight integers is $8$; The median of the eight integers is $8$; The unique mode is $8$ (the value $8$ appears more often than any other value); The range (largest minus smallest) is $8$; Answer choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #6 Guess and Check
The question asks for the biggest a single number can be, which is exactly what Tool #14 (Extreme Principle) is built for: to push one number to its maximum, force every other number down to the smallest value the rules still allow. Tool #4 (Introduce a Variable) pins the picture down — call the largest $L$, so the smallest is $L-8$ from the range. Tool #6 (Guess and Check) does the final work: with a fixed total of $64$, test the top answer choices and either build a valid collection or show one is impossible. The mode clue is the brake — it stops the small numbers from all being tiny, because $8$ has to appear more than anything else.
Execute — Answer: D
6.SP.B.5 Step 1 Turn the four clues into number facts
- Read each clue as a fact about the eight numbers.
- Mean $8$ with $8$ numbers means the total is $8\times 8=64$.
- Median $8$ means the two middle numbers (4th and 5th in sorted order) average $8$.
- Unique mode $8$ means $8$ shows up more times than any other value.
- Range $8$ means the largest number minus the smallest number is $8$.
💡 Each average, mode, and range word is really a plain statement about the eight numbers, so writing them out turns the puzzle into arithmetic.
6.SP.A.2 Step 2 Aim the largest number as high as possible
- To make the largest number as big as it can be while the eight numbers still add to $64$, every other number should be as small as the rules permit.
- Call the largest number $L$.
- The range clue then fixes the smallest number at $L-8$.
- So the whole collection must sit between $L-8$ and $L$.
💡 A fixed total is like a fixed budget: spending less on the other numbers leaves more room for the one you want to be large.
6.EE.B.5 Step 3 Test the biggest choice, 15
- Suppose the largest were $15$.
- Then the smallest is $15-8=7$, so all eight numbers lie between $7$ and $15$.
- Removing the $15$, the other seven numbers must add to $64-15=49$.
- But each of those seven is at least $7$, and $7\times 7=49$ exactly.
- That forces every one of them to equal $7$, giving the collection $7,7,7,7,7,7,7,15$.
- Now the mode is $7$ and the median is $7$, not $8$.
- So $15$ breaks the clues and is impossible.
💡 When seven numbers each at least $7$ must total exactly $49$, there is no slack left, so they are all pinned to $7$.
6.SP.B.5 Step 4 Build a collection with largest 14
- Now try a largest value of $14$; then the smallest is $14-8=6$.
- Give $8$ the most appearances so it is the unique mode: use four $8$'s.
- Fill the rest with three $6$'s and the $14$: the collection is $6,6,6,8,8,8,8,14$.
- Check every clue: the sum is $6+6+6+8+8+8+8+14=64$ so the mean is $8$; the 4th and 5th numbers are both $8$ so the median is $8$; $8$ appears four times while $6$ appears only three, so the unique mode is $8$; and $14-6=8$ is the range.
- Every condition holds.
💡 Loading up on $8$'s secures the mode, and using the smallest legal values for the rest leaves exactly enough total for a $14$.
6.SP.B.5 Step 5 Compare and conclude
- A largest value of $15$ was shown to be impossible, while a largest value of $14$ was built and passes every check.
- Since $14$ works and nothing bigger can, the largest integer that can appear in the collection is $14$, which is choice (D).
💡 The greatest possible value is the largest one you can actually construct without breaking any rule.
6.SP.B.5 Read each clue as a fact about the eight numbers. Mean $8$ with $8$ numbers mean 6.SP.A.2 To make the largest number as big as it can be while the eight numbers still add 6.EE.B.5 Suppose the largest were $15$. Then the smallest is $15-8=7$, so all eight numbe 6.SP.B.5 Now try a largest value of $14$; then the smallest is $14-8=6$. Give $8$ the mos 6.SP.B.5 A largest value of $15$ was shown to be impossible, while a largest value of $14 Review
Reasonableness: The construction $6,6,6,8,8,8,8,14$ is a concrete collection meeting all four conditions, so $14$ is definitely achievable. And $15$ was ruled out cleanly: its seven non-largest numbers would each have to be at least $7$ yet total only $49$, forcing them all to $7$ and wrecking both the mode and the median. So $14$ is both reachable and the ceiling — the answer sits right at the top choice that survives, which is a natural place for a maximization answer.
Alternative: Set up an inequality instead of testing choices. Let the largest be $L$ and put two $8$'s in the middle for the median; to keep the mode $8$ and the smallest at $L-8$, the four smallest values are at least $L-8$. Writing the total as $64$ and demanding the small numbers stay at their minimum leads to $4(L-8)+2\cdot 8+ \ldots \le 64$, which caps $L$ at $14$. Building $6,6,6,8,8,8,8,14$ then confirms the cap is met.
CCSS standards used (min grade 6)
6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Reading the mean, median, unique mode, and range as facts about the eight numbers, and verifying that the collection $6,6,6,8,8,8,8,14$ meets all four.)6.SP.A.2Understand that a set of data collected has a distribution with center and spread (Using the range (a measure of spread) to tie the smallest number to the largest, so pushing the largest up drives the setup.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Arguing that seven numbers each at least $7$ and summing to exactly $49$ must all equal $7$, ruling out a largest value of $15$.)
⭐ To make one number as big as possible under a fixed total, squeeze every other number down to the smallest the rules allow — then check the biggest choice actually builds.
⭐ To make one number as big as possible under a fixed total, squeeze every other number down to the smallest the rules allow — then check the biggest choice actually builds.
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