Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #9
Grade 6 arithmetic
Pick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only 6 pairs, so the most direct attack is Tool #2 (Systematic List): list the pairs in order (1,7), (2,8), …, (6,12), compute each average, and then average those 6 numbers. While doing the list we can use Tool #5 (Pattern): the six pair-averages turn out to be 4, 5, 6, 7, 8, 9 — a perfectly regular arithmetic run, whose average is just the midpoint (4+9)/2. Tool #1 (Diagram) shows why every pair must sum to 1+7 = 13, 2+8 = 10, 3+9 = 12 … wait, they do NOT all share the same sum, but they do all share the same average of (a+b)/2 that creeps up by 1 each time — exactly the pattern we exploit.
Pair up opposite numbers
Pair each clock number with the one across — 'opposite' means 6 positions apart, so k pairs with k+6.
Generating opposite pairs by the rule "k pairs with k+6" is a Grade 4 "follow a given rule to make a pattern" move.
4.OA.C.5Make A Systematic ListAverage each pair
Average each pair with , which gives the list 4, 5, 6, 7, 8, 9.
Each pair sum is at most 18 and dividing by 2 is fluent Grade 3 multiplication/division within 100.
3.OA.C.7Make A Systematic ListAverage the six results
These six averages are consecutive integers, so their mean is just the midpoint — the average of the first and last.
Recognizing that the average of an evenly spaced list equals its midpoint is a Grade 6 "summarize a numerical data set with a measure of center" insight.
For the six pair-averages 4, 5, 6, 7, 8, 9 — a run that climbs by the same step each time — the average of the whole run equals its midpoint, the average of the first value and the last, (4+9)/2.
▸ Why?
The average is the total of the numbers divided by how many there are, so once the total is shown to equal the count times the midpoint, dividing by the count must leave exactly the midpoint.
▸ Why?
Dividing a product by one of the two factors that made it gives back the other factor untouched, so the count times the midpoint, divided by the count, returns the midpoint.
▸ Why?
The total really does equal the count times the midpoint: fold the run end to end into pairs — smallest with largest, next-smallest with next-largest — and every pair adds to the same amount, twice the midpoint, so all the pairs together come to the count times the midpoint.
▸ Why?
Each end-to-end pair adds to the very same total, because stepping one place inward the lower number rises by the fixed step exactly as the upper number falls by that same step, and the rise and the fall reverse each other so the pair's sum never changes.
▸ Why?
Adding up those equal pair-totals gives the run's own total, because the pairs between them use every number exactly once, with none skipped and none counted twice.
Check by adding and dividing
Confirm by adding all six values to get 39, then dividing by 6 to reach the same value.
Reading 39/6 as "39 divided by 6" and simplifying to 6.5 is Grade 5 "fraction as division".
5.NF.B.3Make A Systematic ListThis AMC 8 problem only needs Grade 6 "average is the center of the data" reasoning you already know — and once you see the pair-averages line up as 4, 5, 6, 7, 8, 9, the answer is just the middle: 6.5!
- Pair up opposite numbers
- Average each pair
- Average the six results
- Check by adding and dividing
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