Problem 13 · 2015 AMC 8
Medium
Counting & Probability
sum-mean-relationshippair-counting
How many subsets of two elements can be removed from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} so that the mean (average) of the remaining numbers is 6?
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Answer: D — 5 pairs.
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Hint 1 of 2
'Mean of the leftovers is 6' sounds vague, but a mean is just a total in disguise: with a known count, fixing the mean fixes the sum. So this secretly tells you the exact total the removed pair must carry away.
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Hint 2 of 2
Compute the starting total and the required ending total; their difference is the sum the two removed numbers must hit. Then it's pure pair-counting.
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Approach: convert the mean into a fixed sum (sum = mean × count)
- Whole-set sum: 1 + 2 + … + 11 = 66. Removing 2 leaves 9 numbers, and mean 6 forces their sum to be 9 × 6 = 54.
- So the two removed numbers must sum to 66 − 54 = 12 — that single number is the whole puzzle now.
- Count pairs from {1, …, 11} adding to 12: {1,11}, {2,10}, {3,9}, {4,8}, {5,7} — pairing inward from the ends until they'd cross at 6. That's 5 pairs.
- Why this transfers: whenever a problem fixes an average, immediately rewrite it as a sum — sums add and subtract cleanly, averages don't.
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