AMC 10 Step by Step

2016 AMC 10A Problem 8

P1-10~2 minPrint

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?

Log in to save your progress.
Show solution without answering

Key insight

Work backwards from 0 coins. To undo a crossing, add back the 40 coins and then cut the amount in half. Three undo steps give 20, then 30, then 35.

Solution

We know how the story ends (Fox has 00 coins), so run it backwards.

1. One crossing does two things in this order: double the money, then pay 4040.
2. To undo a crossing, undo those two things in reverse order: first add the 4040 back, then halve.
3. After the third crossing Fox has 00.
- Before the third crossing: (0+40)÷2=20(0 + 40) \div 2 = 20.
- Before the second crossing: (20+40)÷2=30(20 + 40) \div 2 = 30.
- Before the first crossing: (30+40)÷2=35(30 + 40) \div 2 = 35.
4. Check forwards: 35→70→3035 \to 70 \to 30, then 30→60→2030 \to 60 \to 20, then 20→40→020 \to 40 \to 0. Fox is broke after three crossings, as the story says.

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

Why this works

When you know the final result and each step can be undone, working backwards replaces algebra with a few easy calculations. The key is to undo the last action first, just as you take off your shoes before your socks. Always check your answer by running the story forwards.

Alternative approach

Use a letter. Start with xx coins.

  • After crossing 1: 2x−402x - 40.
  • After crossing 2: 2(2x−40)−40=4x−1202(2x - 40) - 40 = 4x - 120.
  • After crossing 3: 2(4x−120)−40=8x−2802(4x - 120) - 40 = 8x - 280.

Setting 8x−280=08x - 280 = 0 gives x=35x = 35.

Concise solution

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

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}.

Review in the AoPS books

Similar problems

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