Problem 23 · 2000 AMC 8
Stretch
Arithmetic & Operations
average
There is a list of seven numbers. The average of the first four numbers is 5, and the average of the last four numbers is 8. If the average of all seven numbers is 647, then the number common to both sets of four numbers is
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Answer: B — 6.
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Hint 1 of 2
Averages convert to sums (sum = average × count). The first four and last four together name 8 slots — but there are only 7 numbers, so ONE number sits in both groups and gets counted twice.
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Hint 2 of 2
If (first four) + (last four) counts the shared number twice and everything else once, then subtracting the honest total of all seven (each once) cancels everything except that one extra copy.
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Approach: the shared number is the only thing counted twice
- Turn averages into sums: first four = 4 · 5 = 20, last four = 4 · 8 = 32. Add: 20 + 32 = 52 — this counts the overlapping middle number twice and the other six once each.
- The honest total of all seven is 7 · 6⁴⁄₇ = 7 · (46/7) = 46, counting every number exactly once.
- Subtract: 52 − 46 = 6. The subtraction removes one full copy of every number, leaving precisely the extra copy of the shared one.
- You'll see it again: this is inclusion–exclusion in miniature — when groups overlap, sum-of-groups minus true-total isolates the overlap. Always start by converting averages to sums so they can be added.
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