Problem 16 · 2025 AMC 8
Hard
Number Theory
complementary-countingsum-constraint
Five distinct integers from 1 to 10 are chosen, and five distinct integers from 11 to 20 are chosen. No two numbers differ by exactly 10. What is the sum of the ten chosen numbers?
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Answer: C — 105.
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Hint 1 of 2
The question asks for one fixed sum, yet you get to choose the numbers freely. That's a hint the sum is the same for every valid choice — so find what's forced. Each low number you pick rules out exactly one high number.
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Hint 2 of 2
Picking 5 lows blocks 5 highs, leaving exactly 5 highs you're forced to take. Those forced highs are 10 more than the 5 lows you didn't pick — so chosen-highs and unchosen-lows pair up.
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Approach: the high picks are forced; chosen highs pair with the unchosen lows
- Insight: the answer can't depend on your choices (the problem gives just one sum), so something must be forced. Each chosen low x blocks the high x+10; choosing 5 lows blocks 5 highs, so the 5 highs you must take are precisely the unblocked ones — the highs that are 10 more than the 5 lows you didn't choose.
- So every high pairs with an unchosen low: sum of chosen highs = (sum of unchosen lows) + 5×10. Then (chosen lows) + (chosen highs) = (chosen lows) + (unchosen lows) + 50 = (all of 1–10) + 50.
- 1 + 2 + … + 10 = 55, so the total is 55 + 50 = 105, the same for any legal choice.
- Why this transfers: when a problem lets you choose but asks for a single number, look for the invariant — the quantity that's the same across all valid choices. Here a blocking/pairing argument shows the sum never moves.
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