Competition · AMC preparation · step 4 of 4
AMC 8 · 2015 · #13
Grade 7 arithmeticcountingPick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Add up the whole set
Add up the whole set: the eleven consecutive integers sum to 66.
Pairing 1+11, 2+10, …, 5+7 gives five 12's plus the middle 6, which is 5 · 12 + 6 = 66 — a Grade 4 pattern in arithmetic.
4.OA.B.4Identify SubproblemsFind the sum that must stay
The mean is the sum over the count, so the nine kept numbers must total 54.
Working backwards from "mean = 6 over 9 numbers" pins down the total — this is the Grade 6 statistics definition of mean.
6.SP.B.5Convert To AlgebraSubtract to get the removed sum
Subtract to see the removed pair must carry the difference, a sum of 12.
Total = kept + removed, so removed = total - kept. This is a one-step Grade 4 word-problem subtraction.
The two numbers removed from the set must add up to 12.
▸ Why?
What the removed pair adds to is forced, because the whole set and the kept nine are both pinned down, and the removed part is simply the whole minus the kept.
▸ Why?
The three totals are tied together: the removed pair plus the kept nine is exactly all eleven numbers, so the removed total equals the whole total minus the kept total.
▸ Why?
Splitting the eleven numbers into the removed pair and the kept nine loses nothing and counts nothing twice, so the two group-totals add back to the whole total.
▸ Why?
Reading the removed total out of "whole = removed + kept" means undoing the addition, that is, subtracting the kept total from the whole total.
▸ Why?
Both fixed totals are known: all eleven numbers make 66 and the kept nine must make 54, and 66 minus 54 leaves 12.
▸ Why?
The eleven numbers total 66, checked by pairing 1 with 11, 2 with 10, 3 with 9, 4 with 8, 5 with 7 into equal sums of 12 plus the middle 6 — reordering the addends never changes their total.
▸ Why?
The kept nine numbers must total 54, because their mean of 6 is just their total shared equally among the 9, so the total is that mean taken 9 times: 6 × 9 = 54.
List the pairs summing to 12
List every pair of distinct numbers in the set that adds to 12, walking the smaller value up from 1.
Walking a = 1, 2, 3, … and reading off b = 12 - a is the Grade 5 "generate two related patterns" move.
5.OA.B.3Guess And CheckCount the valid pairs
Reject {6, 6} because a subset needs distinct elements, leaving five valid pairs.
Counting unordered pairs of distinct outcomes that meet a condition is the Grade 7 "compound events / sample-space" idea.
7.SP.C.8Guess And CheckThis AMC 8 problem really comes down to one Grade 6 idea — average = sum ÷ count — plus careful pair-counting you already know from Grade 7.
- Add up the whole set
- Find the sum that must stay
- Subtract to get the removed sum
- List the pairs summing to 12
- Count the valid pairs
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