Math Journey Archive

Math Journeys, Volume 3
Boy, Devil, Fractal, Part 1
"Uncle Bob"

This journey will show that a rich problem can provide much learning over quite a range of mathematical sophistication. "The Boy and the Devil" problem can be solved using simple arithmetic or basic algebra, and, by asking questions upon reflection, it can be used to introduce the topics of function, recursion and even escape-time fractals.

The problem involves a boy with some money, the Devil, and a bridge. The Devil offers the boy a financial enticement. He will double the youth's holdings every time he crosses the bridge. There will be a modest six-dollar toll to pay at the far end. After one crossing, the boy, in fact, has his money doubled and then he pays the toll. Thinking that he must get rich in the long run, the boy crosses twice more under these rules and finds himself flat broke! How much money did he start with?

This journey doesn't begin with a brisk sleigh ride, does it? It's rather more like a slog through a swamp. Where's the trail map? ... the information booth? I don't see any numbers. To solve we must make good use of what little there is to go on. Hmm. Doubled. Pays the [six-dollar] toll. In that order. Oh yes, and 'flat broke' relates to a number.

Solvers have been observed using the following strategies to solve the problem: trial and error, working backwards, writing formulas, and solving an equation.

Trial and Error. Let's have the boy start with \$8. After a crossing he has \$16 and pays \$6, leaving him with \$10. Continuing for two more crossings he ends up with \$22. Too bad he didn't start with \$8. Your assignment is to pick 12 other starting amounts and calculate the results after three or more crossings. If you happen to find an answer that checks, continue until you have made 12 trials. We may find that the incorrect answers hold information as well.

Working Backwards. You must use inverse operations in reverse order. If the lad ended with zero dollars, he must have had 0 + 6 before he paid the last toll, and he must have had \$3 before the final doubling. In order to work backwards you must add 6 and halve the amount, and do this three times. Go ahead and carry this through to the solution. You will have been much more efficient than when guessing, but you will know one fact and not the 12 facts from the trials above.

Formulas. If we let x stand for the starting amount, the sequence of three crossings can be modeled by

2x – 6 = y
2y – 6 = z
2z – 6 = 0.

Algebra. Hmm. That's quite a few variables. We can eliminate some of them. We have a z in the third formula, but the second formula tells what z is equal to, and so we substitute for it.

2(2y – 6) – 6 = 0 [Can you locate z?]

The first formula gives us a replacement for y. We end up with a formula that contains one variable, and it stands for the solution.

2(2(2x – 6) – 6) – 6 = 0.

Looks a little busy, eh? But look closely. The inner set of parentheses shows what's in the boy's pocket after one crossing. Likewise, the outer set shows the pocket contents after two crossings. Wow! Parentheses and pockets! To solve this you perform exactly the steps you used to work the problem backwards. Compare.

 2(2(2x – 6) – 6) – 6 = 0 2(2(2x – 6) – 6)       = 6    2(2x – 6) – 6        = 3    2(2x – 6)              = 9       2x – 6                = 4.5       2x                      = 10.5         x                      = 5.25 broke! add 6 back halve add 6 halve add 6 halve 0 6 3 9 4.5 10.5 5.25

Reflection. Our Boy started with \$5.25. Let's mess with this problem. What's so special about a \$6 toll? Would the answer be any different if the toll were, say, \$0.32 or \$8? Solve these for the starting amount (x), and try one other toll of your choosing. Look for a connection between x and the toll.

In your 12 trials, you probably saw the boy go broke in fewer than three crossings. Solve for the starting amounts, if the boy crosses 2, 4 or 5 times before going flat broke. This is not just 'practice' but practice with a purpose. How is the connection between x and the toll affected by the number of crossings?

If you think of other ways to mess with Boy & Devil, try them and let me know your results. I've got one more way – solve the original problem after replacing the word 'doubling' with the word "tripling." Use \$0.27 and \$5.40 for tolls.

In part 2, we'll compare our results and make more messes in general.

Continue to Part 2

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