Competition · AMC preparation · step 4 of 4
AMC 8 · 2002 · #3
Grade 6 arithmeticPick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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, …) and take the first four. Tool #2 (Make a Systematic List) gives us a strict ordering so we cannot miss a smaller candidate.
List the even numbers
List the positive even integers in order — the first four are the smallest distinct choices.
Grade 4 multiples: positive even numbers are multiples of 2. Listing them in order makes the four smallest obvious.
4.OA.B.4Make A Systematic ListAdd the first four
The first four are 2, 4, 6, 8 — all distinct positive even numbers — and their sum is 20.
Grade 3 addition within 100. Adding the four smallest gives the smallest possible sum.
3.NBT.A.2Make A Systematic ListDivide by 4
Dividing the sum by four gives the average 5 — no other valid set has a smaller sum, so this is the minimum.
Grade 6 mean: sum ÷ count. The smallest sum and a fixed count of 4 give the smallest mean.
Among every allowed choice, taking the four smallest distinct positive even integers — 2, 4, 6, and 8 — makes the average as small as it can possibly be.
▸ Why?
With the count of numbers fixed at four, the average is just the sum shared into four equal parts, so the sum and the average rise and fall together — whichever choice has the smallest sum also has the smallest average.
▸ Why?
No four distinct positive even integers can add to less than 2 + 4 + 6 + 8, so this choice already owns the smallest possible sum.
▸ Why?
A total is nothing more than its parts joined with no gaps or overlaps, so two choices of four numbers can be compared part against part: if every part of one choice is as small as the rules allow, its total is as small as the rules allow.
▸ Why?
Line the positive even integers up in order — 2 ＜ 4 ＜ 6 ＜ 8 ＜ 10 ＜ … — and any distinct even integer we might bring in instead sits past 8, so it outranks every one of 2, 4, 6, 8: it is greater than 8, and since 8 is greater than each of the other three, it is greater than them too. Trading any chosen number for such a larger one can only push the total up.
Smallest average comes from the smallest sum — so reach for the smallest four positive even integers and take their mean.
- List the even numbers
- Add the first four
- Divide by 4
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