**Math Journeys, Volume 3
Boy, Devil, Fractal, Part 2
"Uncle Bob"**

We began a math journey last month by solving one problem. A boy meets a devil at the entrance to a bridge. The devil determines that the boy has some money, and offers to double the boy's money each time he crosses the bridge, as long as the boy pays the $6 toll at the far end. On first crossing the boy's money doubles and he gladly pays the toll, but after three crossings the boy is flat broke. How much did he start with?

The answer, $5.25, was found using three different strategies: trial and error, working backwards, and solving an equation. I'll demonstrate each method by solving for different bridge tolls under the same rules.

**Trial and error.** If the toll is $0.32, let's see if the boy started with 25 cents. Double that is 50, less 32 is 18; double 18 is 36, less 32 is 4 cents; double 4 is 8, but the boy does not have the third toll. The list that tracks these amounts is called an orbit, and for this example the orbit is 25, 50, 18, 36, 4, and ... -28, if you will.

If the boy starts with more than the toll, he will gain on the devil. Try one example for yourself. Our trial shows that he must've begun with more than 25 cents. Let's put 30 cents in the boy's pocket. The orbit will be 30, 60, 28, 56, 24, 48, 16, 32, and 0. This orbit shows the boy crossing 4 times, but if you look from 28 cents onward, you see that 28 is the exact amount that allows 3 crossings. So these trials, whether involving the solution or not, possess information. If you have trials from the original problem in Part 1 (12 different starting amounts for a $6 toll), then hold them for Part 3 of this journey.

**Working backwards.** To work backwards, we must use inverse operations in reverse order, in other words, instead of doubling and subtracting the toll, we add the toll and halve. If the toll were $8, and the boy went broke after 3 crossings, the orbit run in reverse would be 0, 8, 4, 12, 6, 14, and 7. The boy would have begun with $7.

**Solving an Equation.** Let's solve for a third different toll, that of $16. The equation uses a nest of parentheses to show the boy's pocket contents after each crossing. For a toll of $16, the equation is

2(2(2x – 16) – 16) –16 = 0.

We solve it by using the same steps as in working backward and get x = 14. We check the orbit 14, 28, 12, 24, 8, 16, and 0. It checks.

**Looking for a Pattern.** Let's put all the results in a table.

Initial amount ($) | 5.25 | 0.28 | 7.00 | 14.00 |

Bridge toll | 6.00 | 0.32 | 8.00 | 16.00 |

It seems that in every case the starting amount is 7/8 of the toll. This general solution can be verified by replacing 16 with T in the equation above and following the same solving steps.

Why 7/8 of T? Doubling repeatedly will yield powers of 2 and that might explain the denominator 8. Let's keep tweaking.

**Flexing the problem.** Using 32 cents as the toll and varying the number of bridge crossings can be revealing. For one crossing only, The boy would begin with 16 cents, have it doubled, and pay the toll. He would have started with 1/2 the toll. For 2, 3, 4 and 5 crossings, you would find that the boy began with

3/4, 7/8, 15/16 and 31/32 of the toll.

This lends support to the powers of two denominators and reveals something about the numerators. Check these fractions with other tolls.

**Final Question.** If we flex the problem to investigate "tripling" rather than "doubling" will powers of three be in evidence. Try $0.27 and $5.40 for tolls and see. Will there be a pattern in the numerators too?

In part 3 we take this problem up a level by analyzing these orbits, and in the process, we get an understanding of the concepts involved in escape-time fractals.

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