🇺🇸 AMC 8 ⇄ switch contest
1987 AJHSME

Problem 21

Problem 21 · 1987 AJHSME Stretch
Fractions, Decimals & Percents operator-substitution

Suppose n* means 1⁄n, the reciprocal of n. For example, 5* = 1⁄5. How many of the following statements are true?

i) 3* + 6* = 9*
ii) 6* − 4* = 2*
iii) 2* · 6* = 12*
iv) 10* ÷ 2* = 5*
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Answer: C — 2.
Show hints
Hint 1 of 2
The notation tempts you to add and subtract the bottoms (3* + 6* 'looking like' 9*). But reciprocals behave very differently under + and − than under × and ÷ — translate each line into real fractions before judging.
Still stuck? Show hint 2 →
Hint 2 of 2
Multiplying and dividing reciprocals stays clean: 1⁄a · 1⁄b = 1⁄(ab), and (1⁄a) ÷ (1⁄b) = b⁄a. Adding and subtracting them does NOT just combine the bottoms.
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Approach: rewrite each statement as ordinary fractions and check
  1. i) 3* + 6* = 1⁄3 + 1⁄6 = 1⁄2, but 9* = 1⁄9 — false (you can't add the denominators). ii) 6* − 4* = 1⁄6 − 1⁄4 = −1⁄12, not 1⁄2 = 2* — false.
  2. iii) 2* · 6* = 1⁄2 · 1⁄6 = 1⁄12 = 12* — true. iv) 10* ÷ 2* = (1⁄10) ÷ (1⁄2) = 2⁄10 = 1⁄5 = 5* — true.
  3. Exactly 2 statements hold.
  4. Why this transfers: the trap is assuming a tidy multiplication rule (1⁄(ab)) carries over to addition. It doesn't — products and quotients of reciprocals stay reciprocals, but sums and differences need a common denominator.
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