*Jeopardy!*, Problem-Solving Strategies, and Keyboard Puzzles

My father-in-law was a three-day *Jeopardy!* champion in 1967. Some 50 years later, he is still a devotee of the show, and he and his wife watch religiously every evening. It’s not uncommon for them to call us at 7:25 p.m. to share that day’s Final Jeopardy question. One night recently, this is the question they shared:

THE NAME OF THIS U.S. STATE CAN BE TYPED USING LETTERS FROM ONLY ONE ROW OF A KEYBOARD

I like the question well enough, but what really intrigued me was my mother-in-law’s problem-solving strategy. In what could best be described as guess-and-check, she would randomly name a state and then test it. “How about Delaware? Does that work? No, the E is in the top row,” she’d realize. “What about New Jersey? No, that’s not it, either.” And so she continued for several minutes.

My sons, on the other hand, asked to borrow my smartphone. “You can’t just look up the answer,” I told them.

“We’re not going to,” Alex said. “We just need to see what a keyboard looks like.” They weren’t sure which letters were in each row.

They immediately realized that there are no vowels in the bottom row of the keyboard, so that wouldn’t work. They also noticed that there are four vowels in the top row, so that could involve a lot of searching. So they decided to focus on the middle row, whose only vowel is an A.

Are there any states with only A and no other vowels? Yes, in fact, there are four of them: Alabama, Alaska, Arkansas, and Kansas. And maybe there’s a fifth, depending on whether you consider Y as a vowel; if not, then Maryland has only A’s, too.

It’s left as an exercise for the reader to determine which of those is the answer to the Final Jeopardy question.

Here’s a keyboard-related joke:

A math professor asked one of his graduate students to step into his office. “I need someone to type a bunch of letters for me,” the professor said, “so I’m going to give you a test.” The professor then pointed to a desk with a computer on it, handed him an article from a local newspaper, and told the grad student to reproduce the article. The grad student open Microsoft Word but, not wanting to become a secretary for the professor, proceeded to type very slowly, hunting and pecking with one finger at a time, and making deliberate errors. The professor stopped him after a few minutes. “That’s perfect,” said the professor. “Come back tomorrow morning and I’ll give you the assignment.”

“But aren’t you going to check my work?” the grad student asked.

“Nah,” said the professor, smiling. “You’re the first one who didn’t open Mathematica as soon as you sat down.”

And here are some other keyboard-related questions:

- What’s the longest word that can be typed using the letters from only one row of a keyboard?
- What’s the longest word that can be typed using only the left hand?
- What’s the longest word that can be typed using only the right hand?
- Nearly 90% of humans are right-handed, but our left hands do more of the work when using a keyboard. On average, what percent of letters are typed with the left hand?
- What is the third-most used button on a computer keyboard?
- If you type 10,000 words on a QWERTY keyboard, approximately how far will your fingers have traveled?
- Worldwide, approximately how many times is the space bar pressed every second?
- According to Ray Tomlinson, the inventor of email, what was likely the body of the first email message ever sent?

Answers:

*typewriter**stewardesses**polyphony*(half-credit for*lollipop*even though it has one less letter, since it’s far more common than*polyphony*)- 56%
- backspace (behind
*e*and the space bar) - about a mile
- 6,000,000, according to Keyshorts
- QWERTYUIOP, as part of a test email that Tomlinson sent to himself

## Our Library’s Summer Math Contest

Every summer, our local library runs a contest called *The Great Big Brain Game*. Young patrons who solve all of the weekly puzzles receive a prize. The second puzzle for Summer 2017 looked like a typical math competition problem:

Last weekend, the weather was perfect, so you decided to go to Cherry Hill Park. When you got there, you saw that half of Falls Church was at the park, too! In addition to all the people on the playground, there were a total of 13 kids riding bicycles and tricycles. If the total number of wheels was 30, how many tricycles were there?

First, some comments about the problem.

- I dislike using “you” in math problems. I believe it’s a turn-off to students who can’t see themselves in the situation described. There are enough reasons that kids don’t like math. Why give them another reason to shut down by telling them that they went somewhere they didn’t want to go or that they did something they didn’t want to do?
- Word problems are not real-world just because they use a local context, and this one is no exception. This problem attempts to show an application for a system of linear equations, but true real-world problems don’t have all the information neatly packaged like this.
- Wouldn’t the person posing this problem already have access to the information they seek? That is, if she counted the number of kids riding bikes and the total number of wheels, couldn’t she have just counted the number of bicycles and tricycles instead? It has always struck me as strange when the (implied) narrator of a math problem wants you to figure out something they already know.

All that said, this was meant to be a fun puzzle for a summer contest, and I don’t mean to scold the library. I don’t know that I’d use this puzzle in a classroom — at least, not presented exactly like this — but I love that kids in my town have an opportunity to do some math in June, July, and August.

Now, I’ll offer some comments on the solution. In particular, the solution provided by the library was different than the method used by one of my sons. Here’s what the library did:

Imagine that all 13 kids were on bicycles with 2 wheels. That would be a total of 26 wheels. But since 30 wheels are needed, there are 4 extra wheels. If you add each of those extra wheels to a bicycle, that’ll create 4 tricycles, leaving 9 bicycles. So, there must have been 4 tricycles at Cherry Hill Park.

And here’s what my son did:

If you can’t see what he wrote, he created a system of two equations and then solved it:

2a + 3b = 30

a + b = 13a + 2b = 17

13 – b + 2b = 17

b = 42a + 12 = 30

2a = 18

a = 9

That’s all well and good. In fact, it’s perfect if you want to assess my son’s ability to translate a problem and solve a system of equations. But I have to admit, I was a little disappointed. What bums me out is that he went straight to a symbolic algorithm instead of considering alternatives.

I think I know the reason for this. This past year, my son was in a pull-out math program, in which he studied math with someone other than his regular classroom teacher. In this special class, the teacher focused on preparing him to take Algebra II in sixth grade when he enters middle school. Consequently, students in the pull-out class spent the past year learning basic algebra. My fear is that they focused almost exclusively on symbolic manipulation and, as my former boss liked to say, “Algebra teachers are too symbol-minded.”

A key trait of effective problem solvers is flexibility. That type of flexibility comes from solving many problems and filling your toolbox with a variety of strategies. My worry — and this isn’t just a concern for my son, but for every math student in the country — is that students learn algorithms at the expense of more useful problem-solving heuristics. What happens when my son is presented with a problem that can’t be translated into a system of linear equations? Will he know what to do when he doesn’t know what to do?

The previous pull-out teacher said that when she presented my sons with problems that they didn’t know how to solve, their eyes would light up. They liked the challenge of doing something they hadn’t done before. I’m hopeful that this enthusiasm isn’t lost as they proceed to higher levels of mathematics.

## Why is Today “Prime Day”?

Today is July 11, which the marketing folks at Amazon* have dubbed “Prime Day.” They’ve even created a spiffy, little banner image for it:

Ooh… pretty!

The selection of 7/11 as Prime Day was no doubt deliberate, since both 7 and 11 are prime numbers, though one has to wonder why Amazon ignored the other 52 (or 53, if it’s a leap year) dates they could have chosen:

- February
- 2/2
- 2/3
- 2/5
- 2/7
- 2/11
- 2/13
- 2/17
- 2/19
- 2/23
- 2/29 (some years)

- March
- 3/2
- 3/3
- 3/5
- 3/7
- 3/11
- 3/13
- 3/17
- 3/19
- 3/23
- 3/29
- 3/31

- May
- 5/2
- 5/3
- 5/5
- 5/7
- 5/11
- 5/13
- 5/17
- 5/19
- 5/23
- 5/29
- 5/31

- July
- 7/2
- 7/3
- 7/5
- 7/7
- 7/11
- 7/13
- 7/17
- 7/19
- 7/23
- 7/29
- 7/31

- November
- 11/2
- 11/3
- 11/5
- 11/7
- 11/11
- 11/13
- 11/17
- 11/19
- 11/23
- 11/29

One of my favorite problems is based on the numbers 7 and 11. Here’s a modified version of it, tailored to Amazon’s special day:

An online shopper placed four items in his cart. When he checked out, his credit card was charged $7.11. Shortly thereafter, a programmer realized there was an error in the code, and total price had been calculated by

multiplyingthe prices of the four items. The customer service department was about to alert the customer to the error, but the programmer informed them that the total price would have still been $7.11 if the prices had beenadded. No harm, no foul.There was no sales tax. What was the cost of each item?

Good luck! Happy shopping!

* No, Amazon did not pay me to write a blog post about Prime Day.

## Why Joe Doesn’t Think He’s Average

Today is an average day, with 182 days left in the year, and 182 days in the rear-view mirror.

I know a lot of average jokes:

When my stats teacher said that I was average, she was just being mean.

With my head in a fire

And my feet on some ice,

I’d say that, on average,

I feel rather nice.Two men are sitting in a bar when Mark Zuckerberg walks in. One of the men says to his friend, “How awesome! On average, everyone in this bar is a billionaire!”

The last joke highlights the issue with using the arithmetic mean when the distribution would be more meaningful. Let’s assume that one person in the bar has a net worth less than $10,000, two people have net worth between $10,000 and $100,000, and nine people have net worth between $100,000 and $1,000,000. (These are reasonable estimates for the distribution of net worth in the U.S., by the way.) Then a hypothetical histogram with a logarithmic scale showing the net worth of all the people in the bar might look something like this:

The average net worth of all 13 people in the bar is over $1,000,000,000 — actually, it’s over $4,000,000,000, because Zuckerberg’s net worth is around $60 billion — but only one of them actually has that much money.

In general, describing data with its average is a terrible idea…

If you’re an “average” person, then you’re a 5’9” (10%) male (50%) with brown eyes (55%) and straight (55%), black hair (85%) who wears size 10.5 (US) shoes (20%). You have 25 teeth in your mouth (30%) — including 4 wisdom teeth (80%) — normal color vision (95%), and O+ blood (40%), but you don’t have dimples (75%). You don’t have hitchhiker’s thumb (75%) or a bent little finger (95%), either, but you can roll your tongue (80%). You have an innie belly button (90%), loops in your fingerprints (65%) instead of whorls or arches, and attached earlobes (65%), and when you interlace your fingers, your left thumb rests atop your right thumb (55%). You sleep 6.5 hours per night (15%), smoke 800 cigarettes per year (10%), and consume 2 alcoholic drinks per week (30%). You’re 29 years old (2%), eat 3 servings of fruits and vegetables a day (20%), get 70 minutes of cardio exercise per week (25%), and have a body mass index (BMI) of 25 (15%).

And thanks to the artistic styling of Paul Wrangles at Sparky Teaching, the average person might look a little something like this:

The numbers in parentheses represent the percent of the world population that has the given characteristic. Admittedly, they’re WAGs; I grabbed each statistic from a random location on the web, and I have absolutely no data to back up any of these claims. Moreover, they’re not very precise; I rounded each to the nearest five percent, because a greater level of precision might give the appearance that they’re somehow more accurate.

That said, I don’t think they’re horribly wrong, either, and even if they’re slightly off, they’ll still serve my point. Which is this: Though this description captures an “average” person, it’s pretty far from representing a typical person. The probability that such a person actually exists is only about 1 in 3,500,000,000.

So if you read the description above and thought, “Hey, that’s me!” then you should feel pretty special, indeed — there is likely only one other person in the world with those same characteristics.

The characteristics for the average person used above are sometimes the **mean** for the category (height, shoe size) and sometimes the **mode** (eye color, fingerprints). Both of these measures of central tendency are known as averages, as is the **median** (which I used at the start of this post when claiming that today is an average day).

Life expectancy is another one of those situations where the average provides misleading — or, at least incomplete — information.

Today, the infant mortality rate worldwide is just under 5%, and life expectancy is 71 years. A very simplistic model for this data is to assume that 19 out of 20 people will live to age 75, but 1 out of 20 will die during their first year of life. This model is clearly wrong, but as George Box said, “All models are wrong, but some are useful.” Check out the math with this model:

This model is useful, because it shows that the 95% of people who survive infancy can expect to live to age 75.

Now compare that to the middle ages, when the infant mortality rate was a staggering 30%, and life expectancy was 35 years. Again using a simplistic model, 7 out of 10 people would live to age 50, while 3 out of 10 would die before they reached the age of 1. The math looks like this:

So, there’s a problem with using the average to talk about life expectancy, because the distribution in the middle ages was badly skewed by so many childhood deaths.

If we compare life expectancy now to the middle ages using the average of the entire population, it’s a distorted picture. But when we remove the deaths as a result of infant mortality, it’s a little less bleak: those living past age 1 today have a life expectancy of 75 years; those living past age 1 in the middle ages had a life expectancy of 50 years. The scales are still tipped heavily in our favor, but it doesn’t seem quite as drastic as a ratio of 71 to 35.

To put this in perspective, the life expectancy in 1950 was just under 50 years. Most of the increase in life expectancy has actually happened in the last century; during the last 70 years, longevity has increased by more than 20 years.

How typical are you? How long will you live? I have no idea, but I do know this: Half of the people you know are below average.

## Friday Word Puzzle

Sometimes, a small word is contained in a longer word. For example, you can see the three-letter word *rid* tucked nicely inside *F riday* in the title for this post, and

*zip*can be found in the middle of

*mar*.

**zip**anSome folks have told me that the following word-in-a-word is particularly appropriate for this blog…

…since my puns put the UGH in LAUGHTER.

Words within words are the basis of today’s puzzle.

Complete each of the nine words below by placing a three-letter word in the blank. The three-letter words that you use all belong to the same category. But there is a tenth three-letter word from the same category that is not used below. What is the category, and what is the missing word?

- OB _ _ _ D
- C _ _ _ PY
- M _ _ _ OT
- PH _ _ _ M
- H _ _ _ SE
- VE _ _ _ D
- CA _ _ _ OU
- EL _ _ _ SE
- LE _ _ _ E

When I started to create this puzzle, I was hoping to give you a similar list in which a short math word was found in a longer word. I found several, but they seem pretty darn hard, and the missing words aren’t always obviously mathy. But for fun, you can try your hand at these, too…

- D _ _ _ Y
- AS _ _ _ E
- SE _ _ _ H
- BU _ _ _ _ SS
- EL _ _ _ _ TH
- C _ _ _ _ RA
- RU _ _ _ _ NT
- SH _ _ _ _ BLE
- BRA _ _ _ _ ILD
- HU _ _ _ _ D
- PR _ _ _ _ _ IFY
- RE _ _ _ _ NT
- WA _ _ _ _ ELON
- DE _ _ _ _ OR
- HO _ _ _ _ SS
- H _ _ _ _ HOG
- IM _ _ _ _ ST

ANSWERS

- OB eye D
- C hip PY
- M arm OT
- PH leg M
- H ear SE
- VE toe D
- CA rib OU
- EL lip SE
- LE gum E

The three-letter words are all parts of the body. The tenth word in that category is *jaw*, which never appears in the interior of a longer word (only at the beginning or end, such as * jawbone* or

*lock*).

**jaw**- D add Y
- AS sum E
- SE arc H
- BU sine SS
- EL even TH
- C hole RA
- RU dime NT
- SH area BLE
- BRA inch ILD
- HU more D
- PR equal IFY
- RE side NT
- WA term ELON
- DE mean OR
- HO line SS
- H edge HOG
- IM mode ST

## The Amazing National Flag of Nepal

The National Flag of Nepal is unique.

Here are several trivia questions about the Nepalese flag:

- It is the only non-quadrilateral national flag in the world. What is its shape?
- It is one of only three national flags where the height is not less than the length. What are the other two?
- What is the sum of the three acute interior angles within the flag?

Question 3 may be difficult to answer without knowing more about the exact dimensions of the flag. For help with that, we turn to the Constitution of Nepal, promulgated 20 September 2015, which contains the following geometric description for the construction of the flag:

**SCHEDULE – 1
The method of making the National Flag of Nepal**

- Method of making the shape inside the border
- On the lower portion of a crimson cloth draw a line AB of the required length from left to right.
- From A draw a line AC perpendicular to AB making AC equal to AB plus one third AB. From AC mark off D making the line AD equal to line AB. Join B and D.
- From BD mark off E making BE equal to AB.
- Touching E draw a line FG, starting from the point F on line AC, parallel to AB to the right hand-side. Mark off FG equal to AB.
- Join C and G.

The traditionalist in me wishes that “line segment” were used instead of “line,” or that the overline were used to indicate those segments, and that a few more commas were inserted to make it more readable. Consequently, the math editor in me feels compelled to rewrite the directions as follows:

But the American in me — given how many times someone in the United States has tried to legislate the value of π — well, I’m just excited to see accurate mathematics within a government document.

The description continues for another 19 exhilarating steps, explaining how to construct a crescent moon in the top triangle, a twelve-pointed sun in the bottom triangle, and a border around the shape described above. Those steps are omitted here — because you surely get the gist from what’s above — but the following “explanation” that appears below the method is worthy of examination:

The lines HI, RS, FE, ED, JG, OQ, JK and UV are

imaginary. Similarly, the external and internal circles of the sun and the other arcs except the crescent moon are alsoimaginary. These are not shown on the flag.

The entirety of this construction, as any classical geometrician would hope, can be completed with compass and straightedge. I cheated a bit and used Geometer’s SketchPad, with this being the resultant mess:

The rough part was placing *C* so that *AC* = *AB* + 1/3 *AB*. Geometer’s SketchPad could have easily measured *AB*, calculated 4/3 of its length, and then constructed a “circle by center and radius,” but that felt like cheating. Instead, I…

- located
*Q*, which is halfway between*A*and*D*; - constructed circle
*A*with radius*AQ = AP*; - constructed circle
*P*with radius*PD*; - constructed circle
*D*_{1}with radius*DA*; - located the intersections of circle
*P*and circle*D*_{1}at points*X*and*Y*; - constructed a line through
*X*and*Y*; - located
*R*, which is 1/3 of the way from*D*to*A*; - constructed circle
*D*_{2}with radius*DR*; and, - located
*C*, so that*CD*= 1/3*AB*.

Now, you could use that information to determine *CF* and *FG*, and then use the arctan function to calculate the measures of the two acute angles in the upper pennon. If you were so inclined, you’d find that their measures are 32.06° and 57.94°, respectively.

But the question above asked for the **sum** of the three angle measures. Without any work at all, it’s clear that the sum of those two angles must be 90°, since the construction described above implies that Δ*CFG* is a right triangle.

And because *AB* = *AD* by construction, then Δ*DAB* is an isosceles right triangle, and the measure of the third acute angle must be 45°.

And that brings us to a good point for revealing the answers to the three questions from above.

- Pentagon
- The flags of Switzerland and Vatican City are square, so the height and width are equal.
- 135°

If you’re looking for more flag-related fun, check out the MJ4MF post from Flag Day 2016 about converting each flag to a pie graph.

## A Ton of Money (or Maybe More)

One of my favorite resources from Illuminations is Too Big or Too Small, a collection of three classroom activities that develop number sense, one of which is the following problem:

Just as you decide to go to bed one night, the phone rings and a friend offers you a chance to be a millionaire. He tells you he won $2 million in a contest. The money was sent to him in two suitcases, each containing $1 million in one-dollar bills. He will give you one suitcase of money if your mom or dad will drive him to the airport to pick it up. Could your friend be telling you the truth? Can he make you a millionaire?

This problem is from the book *Developing Number Sense in the Middle Grades* by Barbara Reys and Rita Barger, published by NCTM in 1991. So it’s not new, but it’s still good.

My first attempt to use this problem with students was dreadful (details below), but I’ve used this problem successfully many times since. Yet something about it always bothered me. I’m not opposed to fictitious scenarios if they get students interested. But this scenario, in which a friend claims to win $2,000,000 and needs a ride to the airport, seems too contrived and not adventurous enough. Luckily, I recently had food poisoning and spent an entire Saturday on the couch watching bad movies. While watching *Rush Hour* (1998), I found a scenario that I liked a whole lot better…

In the clip, the kidnapper asks for the following:

- $20 million in $50’s
- $20 million in $20’s
- $10 million in $10’s

Now the questions of “How much would that weigh? How big a case would you need to carry all of it?” seem a little more meaningful.

I’ll channel my inner Andrew Stadel here. For both the weight and volume:

- Give an estimate that you know is too low.
- Give an estimate that you know is too high.

Now, do the calculations, and see how close your intuition was.

When I first used this task with students, I was anticipating a great discussion about how to estimate the weight and volume of the money. I suspected that some students might estimate that you could fit 5 or 6 bills on a sheet of paper, there are 500 sheets of paper in a ream, a ream weighs about 5 pounds, yada, yada, yada. Instead, one student raised his hand and said:

A dollar bill weighs exactly 1 gram.

I asked how he knew that. “Do you collect money? Are you a numismatist?”

No. That’s how drug dealers measure cocaine. They put a dollar bill on one side of a scale, and they put the cocaine on the other.

“Oh,” I said.

Some days, your students learn something from you. And some days, you learn something from them.

After you estimate the weight and volume, check your answer by clicking over to reference.com.

If you use this video clip and activity in a classroom with students, I’d love to hear how it goes. Please post about your experience in the comments.