AMC 8 · 2002 · #3
Grade 6 arithmeticProblem
What is the smallest possible average of four distinct positive even integers?
Pick an answer.
AMC 8 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: Pick four different positive even integers. What is the smallest average their four values can have?
Givens: Four integers must be chosen; All four must be positive (greater than $0$); All four must be even; All four must be distinct (no repeats); Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Unknowns: The smallest possible value of the average of the four chosen integers
Understand
Restated: Pick four different positive even integers. What is the smallest average their four values can have?
Givens: Four integers must be chosen; All four must be positive (greater than $0$); All four must be even; All four must be distinct (no repeats); Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Plan
Primary tool: #2 Make a Systematic List
Since the average is the sum divided by $4$, the smallest average comes from the smallest sum. To get the smallest sum from four distinct positive even integers, list the positive even integers in order ($2, 4, 6, 8, 10, \ldots$) and take the first four. Tool #2 (Make a Systematic List) gives us a strict ordering so we cannot miss a smaller candidate.
Execute — Answer: C
4.OA.B.4 Step 1 - List the positive even integers in increasing order.
- The first four are the smallest possible distinct choices.
💡 Grade 4 multiples: positive even numbers are multiples of $2$. Listing them in order makes the four smallest obvious.
3.NBT.A.2 Step 2 - Take the first four from the list: $2$, $4$, $6$, $8$.
- These are distinct, positive, and even, so they meet every condition.
- Add them up.
💡 Grade 3 addition within $100$. Adding the four smallest gives the smallest possible sum.
6.SP.B.5 Step 3 - Divide the sum by $4$ to get the average.
- This is the smallest average because any other valid choice would replace one of $\{2,4,6,8\}$ with a larger even number, raising the sum.
💡 Grade 6 mean: sum $\div$ count. The smallest sum and a fixed count of $4$ give the smallest mean.
4.OA.B.4 List the positive even integers in increasing order. The first four are the smal 3.NBT.A.2 Take the first four from the list: $2$, $4$, $6$, $8$. These are distinct, posit 6.SP.B.5 Divide the sum by $4$ to get the average. This is the smallest average because a Review
Reasonableness: Check the answer is achievable and minimal. The set $\{2, 4, 6, 8\}$ gives average $5$ — a real example, so $5$ is reachable. Now try to beat it: any other set of four distinct positive even integers must drop one of these four and pick a different even number that is at least $10$ (the next even integer not in the set). Swapping, say, $8$ for $10$ raises the sum from $20$ to $22$, raising the average to $5.5$. No swap can lower the sum, so $5$ is indeed the smallest possible average.
Alternative: Tool #6 (Guess and Check): Try the cheaper-looking answers first. Average $3$ would need a sum of $12$ from four distinct positive even integers — but $2+4+6 = 12$ already uses up the budget with only three numbers, leaving nothing for a fourth distinct positive even integer. Average $4$ would need a sum of $16$ — the smallest four-term sum is $2+4+6+8 = 20 > 16$, so $4$ is also impossible. Average $5$ matches $2+4+6+8 = 20$ exactly, so (C) is the smallest reachable average.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs for a whole number and recognize multiples (Recognizing that the positive even integers are the positive multiples of $2$ so the list $2, 4, 6, 8, \ldots$ is complete and ordered.)3.NBT.A.2Fluently add and subtract within 1000 (Adding the four smallest positive even integers, $2 + 4 + 6 + 8 = 20$.)6.SP.B.5Summarize numerical data sets, including reporting the number of observations and measures of center (Using the definition of mean (sum $\div$ count) to convert the smallest sum $20$ into the smallest average $\tfrac{20}{4} = 5$.)
⭐ Smallest average comes from the smallest sum — so reach for the smallest four positive even integers and take their mean.
⭐ Smallest average comes from the smallest sum — so reach for the smallest four positive even integers and take their mean.
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