🇺🇸 AMC 8 ⇄ switch contest
1992 AJHSME

Problem 1

Problem 1 · 1992 AJHSME Medium
Arithmetic & Operations pair-terms
10 − 9 + 8 − 7 + 6 − 5 + 4 − 3 + 2 − 11 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + 9=
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Answer: B — 1.
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Hint 1 of 3
Before reaching for the long addition, notice the alternating + and − signs. What happens if you group each plus-then-minus into a pair?
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Hint 2 of 3
Pairing adjacent terms turns a scary 10-term sum into a count of identical chunks — a trick for any alternating +/− list.
Still stuck? Show hint 3 →
Hint 3 of 3
Each pair like (10 − 9), (8 − 7), … equals exactly 1, so the top is just "how many pairs?"
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Approach: pair off the alternating signs so each pair collapses to 1
  1. The signs alternate +, −, +, −… so group them: top = (10 − 9) + (8 − 7) + (6 − 5) + (4 − 3) + (2 − 1). Each pair is 1, and there are 5 pairs, so the top is 5.
  2. The bottom starts with a minus: (1 − 2) + (3 − 4) + (5 − 6) + (7 − 8) + 9 = (−1)(4) + 9 = 5.
  3. Both equal 5, so the fraction is 5 ÷ 5 = 1.
  4. Why this transfers: any time terms alternate + and −, pairing neighbours turns the whole sum into "number of pairs × pair-value" — far safer than adding ten signed numbers one at a time.
  5. Sanity check: top and bottom use the same digits 1–9 (the top also has a 10), so it's no surprise they land close; equal-and-equal makes the answer exactly 1, not a messy decimal.
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