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2016 AMC 10A Problem 8

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2016 AMC 10A Problem 8AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 2 of 5 · about 2 min · position P1-10

Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays 4040 coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?

  1. A)

    2020

  2. B)

    3030

  3. C)

    3535

  4. D)

    4040

  5. E)

    4545

Answer

C

Key insight

Reverse each crossing: add the 40-coin toll back, then halve; three reversals from 0 give 20, 30, 35.

Solution

Each crossing sends an amount mm to 2m−402m - 40. Undoing a crossing means adding 4040 and then halving.

Fox ends with 00 coins after the third crossing.

  • Before the third crossing: (0+40)/2=20(0 + 40)/2 = 20.
  • Before the second crossing: (20+40)/2=30(20 + 40)/2 = 30.
  • Before the first crossing: (30+40)/2=35(30 + 40)/2 = 35.

Check forward: 35→70−40=30→60−40=20→40−40=035 \to 70 - 40 = 30 \to 60 - 40 = 20 \to 40 - 40 = 0.

The answer is (C) 35\boxed{\textbf{(C)}\ 35}.

Why this works

When the final state is known and each step is invertible, running the process backwards avoids algebra entirely. The inverse of "double then subtract 4040" is "add 4040 then halve," and the order of the inverse steps is reversed.

Alternative approach

Let the start be xx. After three crossings Fox has 8x−40(4+2+1)=8x−280=08x - 40(4 + 2 + 1) = 8x - 280 = 0, so x=35x = 35. Note the fixed point of m↦2m−40m \mapsto 2m - 40 is 4040: starting below it, Fox's money shrinks each crossing, which is why he ends at zero.

The trap

Reversing in the wrong order (halving before adding 40 back), or running the process forward from a guessed value and giving up.

Common mistakes

  • Reversing in the wrong order (halving before adding 40 back), or running the process forward from a guessed value and giving up.
  • Setting up 8x−120=08x - 120 = 0 by subtracting 4040 three times without accounting for the later doublings of earlier tolls.

Techniques

Start from the end state / desired conclusion and reverse