## Posts tagged ‘solving’

### Two the Hard Way (and an Easy Way)

During our trip to Arizona for winter break, two problems surfaced organically while we were on holiday. (Sorry. Two *math* problems. There were lots of non-math problems, too, but we don’t have time for all that.)

On the plane, the interactive in-flight map showed the outside temp, toggling between Fahrenheit and Celsius. That led to the following MJ4MF original problem, which I thought — and still think — is pretty good:

The conversion between Fahrenheit and Celsius temperatures follows the rule

F= 9/5C+ 32. Sometimes, the temperature is positive for both Celsius and Fahrenheit; sometimes, the temperature is negative for both Celsius and Fahrenheit; and other times, Fahrenheit is positive while Celsius is negative. What is the least possible product of the Fahrenheit temperature and its corresponding Celsius temperature?

You can pause here if you’d like to solve this before I present a spoiler.

Before I reveal two different solutions, allow me to digress. It could be that the statement about the temps sometimes being positive and sometimes being negative is denying a teachable moment. The graph below shows the linear relationship between the two temperature scales. Perhaps a good classroom question is:

When will the product

CFbe positive and when will it be negative?

Or maybe a better question is:

When are

CandFboth positive, when are they both negative, and when do they have different signs?

So, to the problem that I posed. As I thought about it on the plane, I concluded that if *F* = 1.8*C* + 32, then the product *CF* = 1.8*C*^{2} + 32*C*. I then used calculus, found the derivative (*CF*)’ = 3.6*C* + 32, set that equal to 0, and concluded that *C* = ‑8.89, approximately. The corresponding Fahrenheit temperature is *F* = 16, so the minimum product is roughly ‑142.22.

Using calculus was like rolling a pie crust with a steamroller, though. I could have just as easily graphed the parabola and noted its vertex:

If I had used the other form of the rule, namely *C* = 5/9 (*F* ‑ 32), things might have been a little easier. Maybe. In that case, setting the derivative equal to 0 yields *F* = 16, which is arguably a nicer number. But then you still have to find the corresponding Celsius temperature, which is *C* = ‑8.89, and the product is still roughly ‑142.22. So, not much easier, if at all, and again graphing the parabola and noting its vertex would have done the trick:

The only real benefit to using this alternate version of the rule is that it provides a reasonable check. Since both methods — and both graphs — yield an answer of ‑142.22, we can feel confident in the result.

But there’s an easier way to solve this one.

Thinking this was a good problem — and because I like when my sons make me feel stupid — I gave it to Eli and Alex. Within seconds, Eli said, “Well, *F* is positive and *C* is negative between 0°F and 32°F, so the minimum will occur halfway between them at *F* = 16. That means *C* = ‑80/9, so it’s whatever ‑1280/9 reduces to.” (Turns out, -1280/9 = ‑142 2/9 ≈ ‑142.22.)

Eli hasn’t taken calculus, so he doesn’t *know* — or, at least, he hasn’t *learned* — that the minimum product should occur halfway between the *x*– and *y*‑intercepts of the linear graph. Yet, he had an intuitive insight that just happens to be true. As a result, what took me about five minutes of deriving and manipulating took him about five seconds.

The second problem arose at the grocery store. Among our purchases was a box of sugar cubes, which contained, surprisingly, 126 cubes.This number is surprising in the sense that it’s not a number you’ll see very often, except for an occasional appearance in the ninth row of Pascal’s Triangle, or maybe if you’re a chemist searching for stable atoms.

A question that could have arisen from this situation involves surface area and volume:

A rectangular prism with integer dimensions has a volume of 126 cubic units. What is the least possible surface area?

That’s not the question that was shared with Alex and Eli, though. (The answer, if you care, is 162 square units, which results from a 3 × 6 × 7 arrangement — which, in fact, is the exact arrangement of cubes in the box above. I suspect this is not a coincidence.)

The problem that I shared with my sons involved probability:

Imagine that the arrangement of cubes is removed from the box intact, and all six faces of the prism are painted red. Then one of the sugar cubes is selected at random and rolled. What is the probability that the top face of the rolled cube will be red?

The boys made an organized list, as follows:

Painted Faces |
Number of Cubes |

3 | 8 |

2 | 40 |

1 | 58 |

0 | 20 |

Further, the boys reasoned:

- P(cube with 3 red faces, red face lands on top) = 8/126 x 1/2 = 8/252
- P(cube with 2 red faces, red face lands on top) = 40/126 x 1/3 = 40/378
- P(cube with 1 red face, red face lands on top) = 58/126 x 1/6 = 58/756

Therefore,

- (P of getting a red face) = 8/252 + 40/378 + 58/576 = (24 + 80 + 58) / 576 = 162/576 = 9/42

Wow! That seems like a lot of work to get to the answer. Surely there’s an easier way, right?

Indeed, there is.

Notice that the penultimate step yielded the fraction 162/576. The numerator, 162, may look familiar. It’s the answer to the question that wasn’t asked above, the one about the least possible surface area of the prism. That’s no coincidence. In total, there will be 162 faces painted red. And there are 6 × 126 = 576 total faces on all of the sugar cubes (that is, six faces on each cube). This again suggests that the probability of rolling a red face is 162/576.

Did you happen to notice that the volume and surface area use the same digits in a different order? Cool.

So there you have it, two problems, each with two solutions, one easy and one hard. Or as mathematicians might say, one elegant and one common.

It’s typical for problems, especially problems worth solving, to have more than one solution strategy. What’s the trick to finding the elegant solution? Sadly, no such trick exists. Becoming a better problem solver is just like everything else in life; your skills improve with practice and experience. It’s akin to Peter Sagal’s advice in *The Incomplete Book of Running*, where he says, “You want to be a writer? […] Just sit down and write. The more you write, the better a writer you will become. You want to be a runner? Run when you can and where you can. Increase your mileage gradually, and your body will respond and you’ll find yourself running farther and faster than you ever thought possible.” You want to be a problem solver? Then spend your time solving problems. That’s the only way to increase the likelihood that you’ll occasionally stumble on an easy, elegant solution.

And every once in a while, you may even solve a problem faster than your kids.

### One, Tooth, Ree, …

Were a state assessment item writer to get his hands on the puzzle that I discuss below, I believe that this is what the problem might look like:

Fifteen congruent segments labeled A-O are arranged horizontally and vertically to form five squares, with five of the segments shared by two squares each. Which three segments can be removed to leave three congruent squares?

- {A, B, C}
- {A, C, E}
- {B, K, N}
- {D, E, F}
- {G, I, N}
- {H, A, L}
- {I, B, M}
- {K, L, M}

You’ve likely seen this problem before, but in a simpler and more elegant form.

Remove three toothpicks to leave three squares.

Eli shared this puzzle with us the other night at dinner, after hearing it during his math enrichment class that day. (It’s my sincere hope that every family has similar conversations at mealtime.) This was a great problem to discuss at the dinner table, since a collection of manipulatives was close at hand.

As you might expect, Eli and his brother started by removing toothpicks randomly and seeing what happened. But it turns out that random toothpick removal is not a great strategy. There are **15 toothpicks** in the arrangement, so there are **15C3 = 495 ways** to select three of them.

Of course, some of those selections are obviously wrong, such as selecting the three toothpicks in the middle row, or choosing the three highlighted in pink:

But even ignoring the obviously wrong ones, that still leaves a lot of unobviously wrong combinations to consider. Yet I let my sons randomly select toothpicks for a few minutes without intervention. Why? Because of a primary tenet for effective problem solving:

Get dirty.

You can’t get any closer to a solution if you don’t try *something*. Even when you don’t know what to do, **the worst thing to do is nothing**. So, don’t be afraid to get your hands a little dirty. Go on, try something, and see what happens. You might get lucky; but if not, maybe it’ll shed some light.

After a few minutes, they conceded that random toothpick selection was futile. That’s when they modified their approach, leading to another important aspect of problem solving:

Look at the problem from a different perspective.

“Hmm,” said Eli. “That’s a lotta toothpicks.” He thought for a second. “But there are only five squares.” That’s when you could see it click, leading to the most popular problem-solving principle on the planet:

Solve a simpler problem.

There are **5 squares** in the arrangement, and there are only **5C3 = 10 ways** to select three of them. If you first choose three squares, then you can count the number of toothpicks that would need to be removed. For instance, you’d have to remove five toothpicks to leave these three squares:

That doesn’t quite work. But the problem is a whole lot easier if you focus on “leave three squares” rather than “remove three toothpicks.” There are only 10 possibilities to check.

From there, it was just a hop, skip, and jump to the solution.

So, maybe this problem, as well as the million or so other remove-some-toothpicks problems, isn’t very mathematical. It only contains two mathematical elements — basic **counting** and **geometry**. In the vernacular of Norman Webb’s Depth of Knowledge Levels, the kinds of questions that you could ask about this problem — “How many toothpicks are there?”, “How many are you removing?”, “What is a square?” — would be DOK 1.

But they’re fun, and they’re engaging to kids, and they develop problem-solving skills, such as those mentioned above as well as perseverance.

For more problems like this, check out **Simply Science’s collection of 16 toothpick puzzles**.

Speaking of the number 16, here’s my favorite toothpick problem.

Remove 4 toothpicks to leave 4 small triangles.

And finally, because I know you came here looking for jokes, here is the only joke I know that involves toothpicks. It’s rather disgusting. Continue at your own risk.

One night, as a bartender is closing his bar, he hears a knock at the back door. It’s a math graduate student. “Can I have a toothpick?” he asks. The bartender gives him a toothpick and closes the door.

Five minutes later, another knock. “Can I have another toothpick?” asks the student. The bartender gives him another one and closes the door.

Five minutes later, a third knock. “Can I have a straw?” asks the student.

“Sure,” says the bartender, and hands him a straw. “But what’s going on out there?”

“Some lady threw up in the back,” says the grad student, “but all the good stuff is already gone.”