Problem 16 · 1990 AJHSME
Hard
Algebra & Patterns
pair-terms
1990 − 1980 + 1970 − 1960 + … − 20 + 10 =
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Answer: D — 1000.
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Hint 1 of 2
The signs go +, −, +, −… so the terms beg to be grouped two at a time. What does each (big − smaller) pair come out to? They're all the same easy number.
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Hint 2 of 2
Group adjacent terms into +/− pairs so each becomes a constant — then it's just (how many pairs) × (that constant). Just be careful about a term left over at the end.
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Approach: group into +/− pairs so each collapses to a constant
- The alternating signs let you pair: (1990−1980), (1970−1960), …, (30−20). Every pair is exactly 10 — a long sum just became 'count the pairs.'
- The numbers run by tens from 20 up to 1990, that's how many pairs: the tops are 1990, 1970, …, 30 — 99 of them. So the pairs give 99 × 10 = 990.
- One term is left unpaired: the final +10. Add it: 990 + 10 = 1000.
- *Worth keeping:* alternating-sign sums almost always want pairing — group neighbors so each pair becomes a constant, then multiply. Always check whether one lonely term is stranded at the end.
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