AMC 10 Step by Step

2018 AMC 10A Problem 1

P1-10~1 minPrint

What is the value of (((2+1)−1+1)−1+1)−1+1?\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?

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Key insight

A power of -1 just means flip the fraction upside down. Start at the innermost brackets and repeat: flip, then add 1.

Solution

The expression looks scary, but it is the same small job repeated three times.

  1. What does the power −1-1 mean? Raising a number to the power −1-1 flips it upside down. For example, 3−1=133^{-1} = \tfrac13 and (43)−1=34\left(\tfrac43\right)^{-1} = \tfrac34.
  2. Where do we start? Brackets are worked from the inside out, so begin with the innermost one, (2+1)(2+1).
  3. Layer 1. 2+1=32 + 1 = 3. Flip: 13\tfrac13. Add 11: 13+1=43\tfrac13 + 1 = \tfrac43.
  4. Layer 2. Flip 43\tfrac43 to get 34\tfrac34. Add 11: 34+1=74\tfrac34 + 1 = \tfrac74.
  5. Layer 3. Flip 74\tfrac74 to get 47\tfrac47. Add 11: 47+1=117\tfrac47 + 1 = \tfrac{11}{7}.

There are no brackets left, so we are done.

The answer is (B) 117\boxed{\textbf{(B)}\ \frac{11}{7}}.

Why this works

Every layer does the same two things: flip the fraction, then add 11. Keeping the result as one simple fraction after each layer means you only ever flip a fraction like 43\tfrac43, never a messy sum. Notice the pattern too: 43→74→117\tfrac43 \to \tfrac74 \to \tfrac{11}{7}, where each new top is the old top plus the old bottom.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Work from the inside out. Each step is "take the reciprocal, then add 11."

  • Innermost: (2+1)−1=13(2+1)^{-1} = \tfrac13, and 13+1=43\tfrac13 + 1 = \tfrac43.
  • Next layer: (43)−1=34\left(\tfrac43\right)^{-1} = \tfrac34, and 34+1=74\tfrac34 + 1 = \tfrac74.
  • Outer layer: (74)−1=47\left(\tfrac74\right)^{-1} = \tfrac47, and 47+1=117\tfrac47 + 1 = \tfrac{11}{7}.

The value is (B) 117\boxed{\textbf{(B)}\ \frac{11}{7}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.