## 2 Good 2 Be True

I was eating a bowl of shepherd’s pie at the Irish pub in our neighborhood. A man walks up to my table and asks, “What’s your favorite number?”

“Uh, 153,” I respond.

“And 153 × 2 is 306,” he says, then hurriedly scurries away.

He approaches another table, asks another patron for her favorite number, and again multiplies it by 2. He does this over and over, popping from table to table, annoying customer after customer. Eventually, the manager notices this eccentric behavior and approaches the man.

“Sir,” says the manager, “You can’t keep interrupting people’s dinners by asking them for a number and then multiplying by 2.”

“What can I say,” he responds. “I love Dublin!”

A little while later, the gentleman at the table next to me says to his companion, “I know a sure-fire way to double your money.”

This piqued my interest, so I leaned over to eavesdrop on his advice.

“Fold it in half,” he said.

Dismayed that you’ve read this far and have only heard two terrible jokes? Well, buck up, because your fortune is about to change. I can’t help you double your money, but I can help you get twice as much for it.

Perhaps you’ve been wanting a copy of **Math Jokes 4 Mathy Folks**, but just haven’t pulled the trigger yet. Well, now’s the time. Robert D. Reed Publishers is offering a BOGO special for MJ4MF, so now you can buy a copy for yourself at regular price and get another for the special math geek in your life **at no charge**!

**http://rdrpublishers.com/blogs/news/yes-math-is-fun**

And check this out.

- If you buy 2 copies, you’ll get 2 additional copies absolutely free!
- If you buy 3 copies, you’ll get 3 more at no cost!
- Buy 4 copies, and 8 copies will be delivered to your door!
- And if you buy 50 copies? Why, you’ll have 100 copies arrive to your home, office, or post office box for the exact same price!
- If you want
*n*copies, you’ll only pay for*n*/2 of them!

Folks, this is a linear relationship that you’d be foolish to ignore!

## Math Problem for 2017

Happy New Year! Welcome to 2017.

Here are some interesting facts about the number 2017:

- It’s prime. (Okay, so that fact isn’t very interesting. But just you wait…)
- Insert a 7 between any two digits of 2017, and the result is still a prime number. That is, 27,017, 20,717, and 20,177 are all prime. (See? Told you it was gonna get better.)
- The cube root of 2017 is approximately 12.63480759, which uses all ten digits 0‑9, and 2017 is the least positive integer that has this property. (Mind blown yet?)
- The decimal expansion of 2017
^{2017}has 6,666 digits. - 2017 = 44
^{2}+ 9^{2} - 2017 = 12
^{3}+ 6^{3}+ 4^{3}+ 2^{3}+ 1^{3}= 10^{3}+ 9^{3}+ 6^{3}+ 4^{3}+ 2^{3}

If you need some more, check out Matt Parker’s video.

Sorry, no video from me. But in honor of our newly minted prime year, I have created a problem for you to solve.

In the area model below (not to scale), the area of the five blue regions is indicated by the number inside the rectangle. What is the area of the yellow region with the question mark inside?

Sorry, I don’t give answers. Feel free to have at it in the comments.

## Predicting Your Child’s Height, and the Eventual Demise of Our Species

I’m between 6’1” and 6’4”, depending on which convenience store I’m leaving. My wife, on the other hand, doesn’t frequent convenience stores, because she can barely see over the counter, not to mention her extreme disdain for Slurpees.

The discrepancy between our heights has its pros and cons. On the plus side, I can reach the pasta pot on the top shelf in the kitchen, and she barely has to bend over to retrieve the Tupperware containers from the bottom shelf. On the down side, we both strain our necks when kissing unless she’s standing on the second step.

Moreover, our sons were measured at 33” tall at their two-year check-ups, which puts them in the 20th percentile. This is when I realized that marrying a woman only 5’2” tall makes it unlikely that either of my boys will earn a spot on the varsity basketball team.

This was a distressing thought. If my sons continued to track at the 20th percentile, they’d reach a height of only 5’7” as adults. When I mentioned this to my mother-in-law, she attempted to allay my fears. “Kids always grow up to be twice their height at age 2,” she told me. That formula predicted their adult height to be 5’6”, which did not reassure me in the least, despite the fact that I didn’t believe it for a second. (Recently, I learned that the Mayo Clinic actually considers this a reasonable formula for predicting height, although they claim that it only works for boys. For girls, they suggest doubling the height at age 18 months.)

Growing up, the shortest person in my family was my oldest sister, who is 5’10” tall. Thinking that my sons might not even reach 5’8” was a major concern for me. As a result, I began to do some research. Although I knew I could not alter the course of their physical development, I had to at least know if they stood a chance at being above average in height.

As it turns out, predicting the adult height of a child is very difficult, and not just for kids whose growth has been stunted by wrestling or gymnastics. I found no fewer than a dozen formulas in widespread use, yet none of them were very reliable. Which is to say, none of them gave the results I was looking for. I spoke with multiple pediatricians who said the best predictor of a child’s eventual height is the mean of the mother’s and father’s heights. That didn’t sit well with me, though, partially because it just seemed too easy, but mostly because it said my sons would be 5’8” tall at adulthood. Still shorter than I’d hoped, so I looked some more.

The Mayo Clinic suggests the following method to predict boys’ height:

- Add the mother’s height and the father’s height.
- Add 5 inches to that sum.
- Divide by 2.

This formula may not be superior to the “double height at age 2” formula above, but at least it felt more mathematically rigorous. Sadly, it predicted that my sons would reach a height of 5’8½“, which is better but still not what I was looking for. So, my search continued.

After several months of research, I finally stumbled on a formula that I thought I could trust. It began like many others, taking the mean of the mother’s and father’s height. But then it compensated for height differences due to gender. For boys, multiply the mean by 13/12; for girls, multiply the mean by 12/13. That is,

Boys:

Girls:

where *m* is the mother’s height and *f* is the father’s height. It roughly means that a boy has an extra inch added to the prediction for every foot in the mean, whereas a girl has an inch subtracted.

Finally, I had stumbled on a formula that predicted my sons would be above average, stretching the tape to just shy of 6’2” tall. This was a win-win, too — this formula predicted that they would be tall, but not quite as tall as their dad.

What’s interesting about this formula is its long-term prognostication. It implies that the human population will continue to increase in size indefinitely. Let me explain.

Google says that the average man is 5’6” tall and the average woman is 5’2” tall. This means that the average human is approximately 5’4”, or 64”, tall. Then an average man and average woman would have an average son who is 69.33” tall and an average daughter who is 59.07” tall — according to the formulas above — and the average human would be 64.20” tall. When this next generation has children, the sons would be 69.55” tall and the daughters would be 59.26” tall, and the average human height would be 64.41” tall. Do you see what’s happening? The average height is increasing slightly with each generation.

It’s not hard to see why this happens. The average of 13/12 and 12/13 is

which means that the average height will increase by 0.3% with each generation. Admittedly, that’s not much, but it portends almost certain doom! By Y3K, the average human will be almost 6’3” tall; and by Y4K, the average human will be nearly 7’3” tall.

There is some historical data to suggest that human height has been increasing for quite some time. In the early 18th century, researchers found that the average English male was 5’5” tall; today, the average English male is 5’9” tall. That’s an increase of about 0.6% per generation. Further, archaeological research shows that the average Greek woman was about 153 cm tall, and the average Greek woman today is closer to 165 cm, an increase of about 0.07% per generation.

Should we expect humans to someday stretch the tape to 8’ or more? Probably not. The paragraph above is a little lesson on how to lie with statistics. While it may be the case that the modern Greek beauty Helena Paparizou (5’7”) is taller than ancient beauty Helen of Troy (5’6”), it hasn’t necessarily been a steady increase. The average female height of a Greek woman in the 10th century was over 162 cm; just two centuries later, the average height dipped below 160 cm.

While the formulas above may serve to make reasonable predictions now, they may not work so well in the future. So if you’re 5’4” tall, don’t be bummed if your great-great-great-great-great-great-grandchild still isn’t able to dunk.

Alex and Eli are now 9 years old, and both of them are 4’3” tall. I strongly suspect that neither of them will reach 6’0”, and that’s just fine. My wife may not be tall, but she is kind, smart, funny, and compassionate, traits that she has generously shared with her children. As a result, my sons are well above average in ways that actually matter.

## It’s Not What’s on the Outside…

Through the Academic and Creative Endeavors (ACE) program at their school, my sons participate in the Math Olympiad for Elementary and Middle School (MOEMS). While passing the door to the ACE room yesterday, I noticed a sign with the names of those who scored a perfect 5 out of 5 on the most recent contest — and my sons’ names were conspicuously absent. Last night at dinner, I asked Alex and Eli what happened, and they told me about the problem that they both missed. (What? Like we’re the only family in America that discusses math problems at the dinner table.) Here’s how they explained it to me:

Nine 1-cm by 1-cm squares are arranged to form a 3 × 3 square, as shown below. The 3 × 3 square is divided into two pieces by cutting along gridlines only. What is the greatest total (combined) perimeter for the two shapes?

*The answer to this problem appears below. Pause here if you’d like to solve it before reaching the spoiler.*

[Ed. Note: I didn’t see the actual exam, so the presentation of the problem above is based entirely on my sons’ description. Apologies to MOEMS for any substantive differences.]

This problem epitomizes what I love about math competitions.

**The answer to the problem is not obvious.**This is the case with many competition problems, unlike the majority of problems that appear in a traditional textbook.**The solution does not rely on rote mechanics.**Again, this differentiates it from a standard textbook problem or — shudder! — from the problems that often populate the databases of many skill-based online programs.**Students have to get messy.**That is, they’ll need to try something, see what happens, then decide if they can improve the result.**Students have to convince themselves when to stop.**Or more precisely, they’ll need to convince themselves that they’ve found the correct answer. For instance, let’s say a student divides the square into a 3 × 2 and a 3 × 1 rectangle. The combined perimeter is 18 units. Is that good enough, or can you do better? This is different from, say, a typical algebra problem, for which students are taught how to check their answer.

The problem also epitomizes what many people hate about math competitions.

**There’s a time limit.**Students have 26 minutes to solve 5 problems. Which means that if students spend more than 5 minutes on this one, they may not have time to finish the other four. (There was a student of John Benson who, when asked about his goal for an upcoming math competition, replied, “I hope to solve half the problems during the competition and all of them by the end of the week.” That’s the way mathematicians work.)**It’s naked math.**Sorry, nothing real-world about this one. (But maybe that’s okay, because real may not be better.)**The problem is presented as a neat little bundle.**This is rarely how mathematics actually works. True problems often don’t present themselves all at once; it’s through investigation and research that the constraints become known and the nuances are revealed.

All that said, *I believe that the pros far outweigh the cons*. Benjamin Franklin Finkel said, “Many dormant minds have been aroused into activity through the mastery of a single problem.” I don’t remember the last time a mind was aroused by the solution to *x* + 7 = -3, but I’ve witnessed awakenings when students solve problems like the one above.

And here’s the tragedy in all of this: **Many teachers believe that only kids who participate in math competitions can handle — or appreciate — math competition questions.** No! Quite the opposite, in fact. Students who have tuned out have done so because they’ve never been challenged and, worse, have never felt the thrill of solving a problem on their own.

What I really love about the problem, though, is it made me think about other questions that could be asked:

- How many ways are there to divide a 3 × 3 square into two pieces that will yield the maximum total perimeter?
- What is the maximum total perimeter if a 3 × 3 square is divided into three pieces?
- What is the maximum total perimeter if a 4 × 4 square is divided into two pieces? …a 5 × 5 square? …a 6 × 6 square?
*The answer to the problem about the 4 × 4 square appears below. Pause here if you’d like to solve it before reaching the spoiler.*

- (wait for it) What is the maximum total perimeter if an
*n*×*n*square is divided into two pieces?

It was that last question that really got the blood pumping.

Here’s a solution for how to divide a 3 × 3 square to yield the greatest total perimeter:

And here’s a solution for how to divide a 4 × 4 square to yield the greatest total perimeter:

What’s the solution for an *n* × *n* square? That’s left as an exercise for the reader.

## NBA, Discovery, and the Math of Basketball

Last week, Discovery Education and the National Basketball Association (NBA) announced a partnership in which real-time data from stats.nba.com will flow into Math Techbook, and students will use that data to solve problems.

How cool is that?

Eighty students from John Hayden Johnson Middle School in Washington, DC, participated in the event, which was emceed by Hall-of-Famer “Big” Bob Lanier and made silly by Washington Wizards mascot G-Wiz.

The event received a lot of press coverage, and as you may have heard, **there’s no such thing as bad publicity**. But one of the articles quoted me as saying:

It’s not like a beautiful, traditional math problem.

That is **not** what I said. I am absolutely certain that I have never used the words *beautiful* and *traditional* in the same sentence. Well, perhaps when referring to a wedding dress or an Irish cottage, maybe, but certainly not when referring to a math problem.

I was also quoted as saying:

It’s going to be messy, for sure.

That is, in fact, one of the things I did say. Because by definition, good math problems are messy. For this project, our writing team created problems that don’t have one right answer. For instance, one problem asks students to **generate a formula to predict which players should be on the All‑NBA 1st Team**. Should they use points and rebounds as part of their formula? If so, how much weight should they give to each? And should there be a deduction for the number of turnovers a player has? All of that is up to the student, and it’s certainly possible that more than one formula would give reasonable results. (If you don’t believe me, do a search for NBA Efficiency, TENDEX, Thibodeau, VORP, or New SPM to get a sense of some formulas currently used by professional statisticians.)

A microsite with a four sample problems is available at **www.discoveryeducation.com/nbamath**. To see all 16 problems and to experience the NBA Math Tool, you’ll need to login to Math Techbook; sign up for a 60‑day trial at **www.discoveryeducation.com/mathtechbook**.

I’m ecstatic about the problems that our writing team — which includes folks who love both math and basketball, like Brenan Bardige, Ellen Clay, Chris Shore, Shauna Hedgepeth, Katie Rhee, Jen Silverman, and Jason Slowbe — has created. One of the simpler problems they’ve written, meant for middle school students, is to **determine which player should take a technical free throw**. It’s not a hard problem, but students get to choose which team(s) to examine and how to use free-throw data to make their choice. With the NBA Math Tool that we’ve created, which includes FTM and FTA but not FT%, one possible formula is **=ROUND(100*FTM/FTA,1)**, which will display the free-throw percentage to one decimal place of accuracy — though there are certainly less sophisticated formulas that will get the job done, too, and students could bypass formulas entirely by using equivalent fractions.

But a different article said that the “questions may look something like” this:

Andrew Wiggins is making 49.1% of his two-point shots and 52.3% of his threes. Which shot is he more likely to make?

Actually, we would **never** ask a question like this in Math Techbook, either as part of this NBA project or otherwise. By the time students start working with percentages in middle school (6.RPA.3.C), we expect that they already understand how to compare decimals (4.NF.C.7). Though basic exercises are included in the service, most problems — and especially those based on NBA data — exist at a greater depth of knowledge.

But what we might do is ask students to use proportions to make a prediction.

As you know, basketball announcers and sportswriters make predictions all the time. They talk about players being “on pace” to score some number of points or to grab a certain number of rebounds. In fact, the Washington Post recently prophesied that Steph Curry will hold the NBA’s all-time three-point record before the next presidential election.

During the first part of the event at Johnson Middle School, the students set out to make a prediction:

How many assists will John Wall finish the season with?

John Wall recently set the Wizards franchise record for assists, so the context was timely.

To solve this problem, students explored the **NBA Math Tool**, which now resides inside Math Techbook. This tool allows students to analyze both NBA and WNBA stats. Students considered data for the Washington Wizards:

Row 6 shows that John Wall had 98 assists through 11 games. Good information, to be sure, but it led to more questions from students than answers:

- Some players on the Wizards have played 13 games. How many games have the Wizards played so far this year?
- How many games will John Wall play this year?
- How many games are in an NBA season?

Looking at team data in the NBA Math Tool, students learned that the Wizards have played 13 games so far this year. And one student knew that every NBA team plays 82 games in a season. Good info… but now what?

One approach is to set up a proportion with the equation

which yields the number of games (*g*) that we can expect John to play this year (69), and then the equation

can be used to find the number of assists (*a*) that we can expect John to record (602).

But the eighty students in the gym were sixth- and seventh-graders, and they weren’t ready for algebraic equations. Instead, they attacked the problem by noting that Wall had 98 assists through the first 13 games, so they estimated:

- He should have about 200 assists through 25 games.
- He’d have about 400 assists through 50 games.
- He’d have about 600 assists through 75 games.
- That’s 7 games shy of a full 82‑game season, and Wall should have about 50 assists in 7 games.
- So, we can expect him to finish the season with about 650 assists.

My role at the event was to lead students through the solution as a group-problem solving activity; and then, to work with them in the media center on the free-throw problem described above. It was an incredible day! I got to co-teach with Ivory Latta, point guard for the Washington Mystics:

I got to meet some incredible people, including current players, former players, and NBA executives:

But most importantly, I was finally able to let the world know about this amazing project, which my team has been working on for a year.

The NBA slogan is,** This is why we play**. But today I say, **this is why we work**: to develop rich curriculum resources that are fun, relevant, and powerful in teaching kids math.

## Fractional Fun

It’s well known that **5 out of 4 people have trouble with fractions**, but even the mathematically advanced may have a little trouble with this puzzle. Your challenge is simple…

Find the sum of all items in the following table.

A hint is below the table, and the answers are below that. Good luck!

Hmm… it seems that you scrolled down here a little too quickly for the hint. Try harder. To put some distance between you and the hint, here are some fraction jokes:

How is sex like a fraction?

It’s improper for the larger one to be on top.Which king invented fractions?

Henry the Eighth.There’s a fine line between the numerator and denominator.

(And it’s called avinculum.)

Okay, you’ve waited long enough. Here’s your hint. The items in the table are a fird (fish + bird), wooden forts, bottles of whiskey, the Sith lord Darth Maul, wraiths, and tents. (By the way, thanks to www.HikingArtist.com for the cool drawing of the fird!) Hope that helps.

To put some space between the hint and the answer, here are some more fraction jokes:

A student once told me, “To prove to you that I understand equivalent fractions, I only did three-sevenths of my homework.”

I was scared half to death… twice.

What is one-fifth of a foot?

A toe.

Okay, you’ve waited (and endured) enough. Without further adieu, the answer is **2.5**. The images in the table are:

- one fird
- two forts
- four fifths
- one Sith
- four wraiths
- two tents

the sum of which is

.

You’re welcome.

Thanks for stopping by. Have a great day!

## Math Millionaire Quiz

It’s hard to believe that *Who Wants to Be a Millionaire* has been on the air since 1999, isn’t it? Even harder to believe is the number of math questions that have been missed by contestants.

In this post, I’m going to share five questions that have appeared on *WWTBAM*, followed by a brief discussion. If you’d like to solve them before reading the discussion, or if you want to share the quiz with friends or students, you can download it:

Three of the five questions were answered incorrectly by contestants. In one case, the contestant polled the audience and received some bad advice. If I hadn’t put this collection together, I’m not sure I would’ve been able to identify which ones were answered correctly. So maybe that’s a bonus question for you: **Which two questions were answered correctly?**

I’m unquestionably biased, but I always feel like the math questions on *WWTBAM* are easier than questions from other disciplines. Then again, maybe a history major would think that questions about Eleanor of Aquitaine are trivial. But take these non-math questions:

- In the children’s book series, where is Paddington Bear originally from? (Wait… he’s not from England?)
- What letter must appear at the beginning of the registration number of all non-military aircraft in the U.S.? (Like most things, it’s obvious — once you know the answer.)
- For ordering his favorite beverages on demand, LBJ had four buttons installed in the Oval Office labeled “coffee,” “tea,” “Coke,” and what? (Hint: the drink wasn’t available when he was Vice President.)

My conjecture is that non-math questions generally have an answer that you either know or don’t know, but math questions can be solved if given enough time to apply some logic and computation.

Perhaps you’ll disagree after attempting these questions.

**1. What is the minimum number of six-packs one would need to buy in order to put “99 bottles of beer on the wall”?**

- 15
- 17
- 19
- 21

**2. Which of these square numbers also happens to be the sum of two smaller square numbers?**

- 16
- 25
- 36
- 49

**3. If a euro is worth $1.50, five euros is worth what?**

- Thirty quarters
- Fifty dimes
- Seventy nickels
- Ninety pennies

**4. How much daylight is there on a day when the sunrise is at 7:14 a.m. and the sunset is at 5:11 p.m.?**

- 9 hours, 3 minutes
- 8 hours, 37 minutes
- 9 hours, 57 minutes
- 8 hours, 7 minutes

**5. In the year she turned 114, the world’s oldest living person, Misao Okawa of Japan, accomplished the rare feat of having lived for how long?**

- 50,000 days
- 10,000 weeks
- 2,000 months
- 1 million hours

**Discussion and Answers**

**1.** Okay, really? Since 16 × 6 = 96, one would need 17 six-packs, **B**.

**2.** This is the one for which the contestant asked the audience. That was a bad move… 50% of the audience chose A, but only 30% chose the correct answer. Since 25 = 9 + 16, and both 9 and 16 are square numbers (9 = 3^{2}, 16 = 4^{2}), the correct answer is **B**.

**3.** It’s pretty easy to calculate $1.50 × 5 = $7.50. The hard part is figuring out which coin combination is also equal to $7.50. Okay, it’s not *that* hard… but it took Patricia Heaton a lifeline and more than 4 minutes.

**4.** My question is whether daylight is officially defined as the time from sunrise to sunset. Apparently, it is. That makes this one rather easy. From 7 a.m. to 5 p.m. is 10 hours, and since 11 and 14 only differ by 3 minutes, we need a time that is 3 minutes less than 10 hours: **C**, final answer.

**5.** Without a doubt, this is the hardest of the five questions. Contestants aren’t allowed to use calculators, so they need to rely on mental math. Estimates will do wonders in this case.

- Days: 114 years × 365.25 days/year ≈ 100 × 400 = 40,000 days
- Weeks: 114 years × 52.18 weeks/year ≈ 120 × 50 = 6,000 weeks
- Months: 114 years × 12 months/year < 120 × 12 = 1,440 months
- Hours: 114 years × 365.25 days × 24 hr/day ≈ 40,000 × 25 = 1,000,000 hours

Only the last result is close enough to be reasonable, so the answer must be **D**.

What’s amusing is that the contestant got the correct answer, but for the wrong reasons. For instance, he estimated the number of weeks to be 50,000, not 5,000. He then used that result to say, “It can’t be 50,000 days, because it’s about 50,000 weeks.” That’s using a false premise to arrive at a correct conclusion. On the other hand, I wonder how well I’d be able to calculate in front of a national audience with $25,000 on the line. Regardless of how he got there, he correctly chose **D**, to which host Terry Crews said, “You took your time on this. You worked it through. It’s what we all need to do in life sometimes. And that’s how you *win the game*!”

Should I ever become a question writer for *Millionaire*, I’d submit the following:

**Which of the following are incorrect answers to this question?**

- B, C, D
- A, B, C
- A, C, D
- A, B, D