πŸ‡ΊπŸ‡Έ AMC 8 ⇄ switch contest
1996 AJHSME

Problem 2

Problem 2 · 1996 AJHSME Easy
Arithmetic & Operations order-of-operations

Jose, Thuy, and Kareem each start with the number 10. Jose subtracts 1 from 10, doubles his answer, and then adds 2. Thuy doubles 10, subtracts 1 from her answer, and then adds 2. Kareem subtracts 1 from 10, adds 2 to his number, and then doubles the result. Who gets the largest final answer?

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Answer: C — Kareem.
Show hints
Hint 1 of 2
All three do the same three operations β€” double, subtract 1, add 2 β€” just in different orders. Notice WHO doubles last. Doubling magnifies whatever you've built up, so saving it for last should help.
Still stuck? Show hint 2 →
Hint 2 of 2
Multiplying last scales up the full amount you've accumulated; doing it early means the later +2 isn't doubled. So predict the person who multiplies last wins β€” then just confirm.
Show solution
Approach: spot who doubles last
  1. Doubling is the only operation that scales the running total, so whoever doubles last gets their biggest number doubled. Kareem is the one who multiplies at the very end β€” so he should win.
  2. Confirm with quick arithmetic: Jose (10 βˆ’ 1)Β·2 + 2 = 20, Thuy 10Β·2 βˆ’ 1 + 2 = 21, Kareem (10 βˆ’ 1 + 2)Β·2 = 22. The largest is Kareem's 22.
  3. Why this transfers: when the same steps run in different orders, ask which step amplifies β€” a Γ—2 or +2 done last lands on a bigger base. The order, not the operations, is the whole game.
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