Competition · AMC preparation · step 4 of 4
AMC 8 · 2000 · #23
Grade 6 arithmeticPick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We never need the seven individual numbers — only their sums. Tool #11 (Find an Invariant) says: the grand total of all seven numbers is the same no matter how you slice them. Tool #9 (Solve an Easier Problem) is the matching move — instead of seven unknowns, turn each average into a sum and reason about three totals: first-four sum, last-four sum, and all-seven sum. Adding the first-four sum and the last-four sum double-counts exactly the overlapping number, so (first-four sum) + (last-four sum) - (all-seven sum) = overlap. No variables, no algebra.
Turn the averages into sums
Turn each average into a sum with sum = average × count — apply it to the first four, the last four, and all seven.
Sums are easier to combine than averages. Convert once and the rest is plain arithmetic.
6.SP.B.5Solve An Easier Related ProblemAdd the two group sums
Add the two group sums to 52 — the shared number sits in both fours, so it gets counted twice.
Two groups of 4 over 7 slots forces exactly one number into both groups. That is the invariant: combined sum = grand total + one extra copy of the overlap.
Adding the first-four total to the last-four total gives the all-seven total plus one extra copy of the number that sits in both groups.
▸ Why?
The first-four total plus the last-four total is the same as taking those four numbers and those four numbers and adding all eight of them together in one sum.
▸ Why?
A group's total is exactly its members added up, so swapping each total for its four numbers keeps the value the same.
▸ Why?
Numbers in a sum can be added in any order, so the eight can be arranged to line them up against the full list one by one.
▸ Why?
In that sum of eight, six numbers show up once and the shared number shows up twice, which is every one of the seven numbers counted once plus a second copy of the shared one.
▸ Why?
The 'first four' and the 'last four' of a seven-long list together reach all seven numbers and share only the one in the middle, so eight slots fall on seven different numbers with exactly one number repeated.
▸ Why?
The all-seven total already counts each of the seven numbers once, with none skipped and none doubled, so it matches the 'seven counted once' part and leaves the extra copy of the shared number behind.
Solve for the overlap
Subtract the all-seven total from the combined sum: 52 - 46 = 6 is the double-counted shared number, choice (B).
The double-counted amount is exactly the gap between the two ways of measuring the same numbers.
6.EE.B.7Work BackwardsConvert the averages into sums: 20, 32, 46. When you add 20 and 32, the shared number gets counted twice — so 52 - 46 = 6 is exactly that shared number. Answer (B).
- Turn the averages into sums
- Add the two group sums
- Solve for the overlap
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