“Doesn’t look much smaller than Djibouti,” Alex replied.

And it’s true. On the map, they don’t look much different in size…

Which led me to wonder, which is larger?

And that one little question led to the creation of a game we now call **More or Less**, in which one person names two items, and the other person needs to identify which of the two is larger. It’s a great game for passing time on a car trip or during a long walk.

Here are some of our favorites:

Category |
Option A |
Option B |

Land Area | Qatar | Djibouti |

Percent of U.S. Flag | Red | White |

Distance from St. John’s, NL | Vancouver, BC | Rome, Italy |

Weight | $10 in Quarters | $10 in Dimes |

Population | New York City | London |

Calories | Big Mac | Whopper |

Equatorial Radius | Neptune | Uranus |

Length | Distance from North to South Pole | Great Wall of China |

Official Capacity | Rungrado May First Stadium, North Korea | Michigan Stadium (“The Big House”) |

Caffeine (per 12 fl. oz.) | Coca-Cola | Pepsi-Cola |

Net Worth | Jeff Bezos | Warren Buffett |

Loudness | Squeeze Toy | Vacuum Cleaner |

Stores Worldwide | Dunkin Donuts | Starbucks |

Number by All Teams in a Season | Home Runs in MLB | Goals in NHL |

Top Speed | Slug | Snail |

Heart Beats per Minute | Pig | Human |

Estimated IQ | Newton | Leibniz |

Number of Factors | 144 | 192 |

Value | 3^{16} |
2^{25} |

You should definitely try to figure out whether Option A or B is larger in each row above, before you look at the answers below.

And if you can answer at least 15 of these correctly, you’re **more or less** a genius!

Answers:

- Maps can be deceiving. Although they look similar in size, the area of Djibouti is more than double that of Qatar. Djibouti is 23,200 km
^{2}, Qatar only 11,586 km^{2}. - The U.S. flag is about 41.5% red, 40.9% white.
- Vancouver is 5,117 km from St. John’s, and Rome is only 5,050 km. (Canada is a big country!)
- A dime weighs 2.268 g, a quarter weighs 5.670 g. So 100 dimes and 40 quarters will both weigh 226.8 g.
- London has 8.7 million people. New York has just slightly fewer with 8.6 million.
- A Whopper (no cheese) has 680 calories, whereas a Big Mac (with its two patties, a slice of cheese, and an extra bun in the middle) has only 540 calories. Go figure.
- Uranus is larger with a radius of 25,500 km; the radius of Neptune is 24,700 km.
- Traveling from the North pole to the South pole would be circumnavigating half the Earth, which is about 12,430 miles. But the Great Wall of China is estimated to be 13,170 miles.
- Rungrado holds 114,000, whereas The Big House only holds just under 108,000.
- Pepsi has 58 mg of caffeine, Coke only 54 mg.
- Jeff Bezos is worth $70 billion, Warren Buffett is worth $65 billion. They’re both ridiculously rich, but Bezos was more efficient in acquiring his wealth.
- A vacuum cleaner will reach 75 dB, which is “slightly annoying,” whereas a squeeze toy can reach 90 dB. A vacuum cleaner seems louder and more annoying because the sound persists, whereas most squeeze toys make a noise once, then stop.
- In 2016, there were 25,085 Starbucks but only 12,258 Dunkin’ Donuts, according to Statista.
- There were 6,672 goals scored in the NHL during the 2015-16 season, but just 5,610 home runs hit in the MLB in 2016. There are more than 6,000 hockey goals scored every year, but only two seasons in the past decade have seen more than 5,000 home runs.
- A fast slug can move 0.2 mph, but the poor snail — with that heavy shell on its back — can only muster about 0.02 mph.
- An average human’s heart beats about 60 times per minute; an average pig’s heart, about 70.
- According to
*The Early Mental Traits of Three Hundred Geniuses*by Catharine M. Cox, Leibniz was 183, Newton 168. (Sorry, Isaac!) - The number 144 has 15 factors, the number 192 has only 14.
- 3
^{16}– 2^{25}= 9,492,289.

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Last week, he emailed me the following:

My garage door opener has an exterior keypad that allows me to open the door by entering a 5‑digit number. There is no ENTER key, so the keypad “listens” for the correct code and disregards a false start. How many key presses would it take to test every possible code?

Theoretically, there are 10^{5} possible codes, so entering all of them sequentially would require 5 × 10^{5} key presses. However — because the keypad ignores false starts — some key presses can be saved. For example, typing 123456 will actually test two codes, 12345 and 23456.

Chris continued by asking:

Is it possible to construct an optimal string of key presses of minimal length that tests every possible code?

And with that, my Tuesday was ruined.

I had seen this problem before, or at least a version of it. The top four students at the MathCounts National Competition compete in a special event called the Masters Round, and one year the problem was about something called **D Sequences**. The author used this nickname because such sequences of minimal length are known as *de Bruijn sequences*, after the mathematician Nicolas Govert de Bruijn who proved a conjecture about the number of binary sequences in 1946.

Luckily for Chris, he caught a nasty viral infection last week, which gave him plenty of time to lie in bed thinking about the problem. He emailed me on Monday to inform me of his progress:

I did not manage to prove anything, but I did write a computer program that generates sequences using a pretty straightforward algorithm, and I was able to confirm that solutions are possible for 2‑, 3‑, 4‑, and 5‑digit codes.

That note reminded me that the best way to ensure a happy life is to surround yourself with intelligent people who share similar interests. Chris concluded his email to me with this:

I’d say [that my garage] is pretty secure, since it would take me about 14 hours to punch in all the possible numbers, reading from a list.

Feel free to read more about **de Bruijn sequences** at MathWorld, but you might want to try the following problems first.

- Construct a de Bruijn sequence that contains every two-digit permutation of 0’s and 1’s.
- Construct a de Bruijn sequence that contains every three-character permutation from an alphabet with three characters.
- What is the minimum length of a string of letters that would contain every possible five-letter “word,” that is, every possible permutation of 5 letters, using the Latin alphabet?

Counting things is something that mathematicians, especially those studying combinatorics, do quite often. Yet how they count can be atypical:

When asked how many legs a sheep has, the mathematician replied, “I see two legs in front, two in back, two on the left, and two on the right. That’s eight total, but I counted every leg twice, so the answer is four.”

And there you have it.

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…and it was delicious!

Appropriate for Pi Day, I suppose, as is the game my sons have been playing…

Eli said to Alex, “18 and 126.”

Alex thought for a second, then replied, “2, 7, and 9.”

“Yes!” Eli exclaimed.

I was confused. “What are you guys doing?” I asked.

“We invented a game,” Eli said. “We give each other **the sum and product of three numbers**, and the other person has to figure out what the numbers are.”

After further inquisition, I learned that it wasn’t just any three numbers but **positive integers** only, that **none can be larger than 15**, and that they must be **distinct**.

Hearing about this game made me immediately think about the famous Ages of Three Children problem:

A woman asks her neighbor the ages of his three children.

“Well,” he says, “the product of their ages is 72.”

“That’s not enough information,” the woman replies.

“The sum of their ages is your house number,” he explains further.

“I still don’t know,” she says.

“I’m sorry,” says the man. “I can’t stay and talk any longer. My eldest child is sick in bed.” He turns to leave.

“Now I know how old they are,” she says.

What are the ages of his children?

You should be able to solve that one on your own. But if you’re not so inclined, you can resort to Wikipedia.

But back to Alex and Eli’s game. It immediately occurred to me that there would likely be some ordered pairs of (sum, product) that wouldn’t correspond to a unique set of numbers. Upon inspection, I found eight of them:

(19, 144)

(20, 90)

(21, 168)

(21, 240)

(23, 360)

(25, 360)

(28, 630)

(30, 840)

My two favorite ordered pairs were:

(24, 240)

(26, 286)

I particularly like the latter one. If you think about it the right way (divisibility rules, anyone?), you’ll solve it in milliseconds.

And the Excel spreadsheet that I created to analyze this game led me to the following problem:

Three distinct positive integers, each less than or equal to 15, are selected at random. What is the most likely product?

Creating that problem was rather satisfying. It was only through looking at the spreadsheet that I would’ve even thought to ask the question. But once I did, I realized that solving it isn’t that tough — there are some likely culprits to be considered, many of which can be eliminated quickly. (The solution is left as an exercise for the reader.)

So, yeah. These are the things that happen in our geeky household. Sure, we bake cookies, play board games, and watch cartoons, but we also listen to the NPR Sunday Puzzle and create math games. You got a problem with that?

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“200?” asked the shepherd. “But we only have 196 sheep.”

The dog replied, “Well, yeah, but you know I like to round up.”

Rounding up has been a topic of conversation in college basketball this week.

Marcus Keene, a guard for the Central Michigan Chippewas, scored 959 points in 32 games this season, giving him a points-per-game (PPG) average of 30.0.

Sort of.

Technically, his average is 29.96875, just shy of the highly coveted 30 points-per-game mark that’s only been attained by a few dozen players in NCAA history. Since 1981, only 8 players have reached 30 PPG, most recently Long Island’s Charles Jones in 1996‑97.

But the controversy swirled this week because Keene didn’t actually average more than 30 points per game. He was one point shy. His lofty accomplishment was nothing more than smoke-and-mirrors due to round-off error, or so the critics say.

Per-game statistics are used to compare players with one another, because totals can’t be compared for players who have played a different number of games. And let’s face it, no one wants to get into the habit of comparing per-game stats to seven decimal places. The NCAA reports all per-game statistics to the nearest tenth, and the truth is that Keene’s PPG average would be reported as 30.0, 30.00, 30.000, and 30.0000 if rounded to tenths, hundredths, thousandths, and ten-thousandths, respectively.

It’s been a good year for math and basketball. Anthony Davis can have an asterisk for his record-setting 52 points in the NBA All-Star Game because no one played defense; and now Marcus Keene can have an asterisk for his 30.0 points-per-game average.

In related news, it was reported that 53% of men say that they will watch the NCAA Division I Men’s Basketball Championship (aka, “March Madness”). And just to prove the men are the dumber sex, 61% of them admitted that they’ll watch while at work. Simple math says that 32.3% of men will watch the tourney at work. Which means that if you’re a man with two friends who don’t like basketball, then you’ll be the one killing office productivity next Thursday.

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And if you want to say, “In fact, you’re so unimportant, I bought your gift card while I was eating breakfast,” then head to a pancake house.

That’s right. Our local pancake shop is now offering gift cards — and, boy, do they have a great sale! Check this out…

You smell it coming, don’t you? No, not a stack of pancakes! I’m referring to the math question on the other side of that image.

**Which is the best deal?**

One way to attack this problem is to compare the free amount to the original price. That is, which fraction is greatest: 2/15, 4/25, 7/50, 11/75, or 15/100?

Ack. Too much work.

A better way is to do a piece-wise comparison:

- Which is better, $15 or $25? Clearly the
**$25**. You get twice as much free for only $10 more. - Which is better, $25 or $50? Well, two $25’s will get you $8 free, but for the same price, you’ll get one $50 card and only $7 free. So, the
**$25**card wins again. - Which is better, $25 or $75? Well, three $25’s will get you $12 free, but for the same price, you get one $75 card and only $11 free. Don’t look now, but
**$25**is on a roll. - Which is better, $25 or $100? Well, four $25’s will get you $16 free, but for the same price, you get one $100 card and only $15 free. There you have it,
**$25**is the champ.

Now that that’s out of the way, you can probably anticipate my next question.

**Who the hell came up with this pricing scheme?**

It’s not typical to get a smaller reward when you spend more money. Usually, the more you spend, the more you get free. Then again, it’s a pancake shop. Maybe they did some significant market analysis, recognized that no one could actually spend $100 on pancakes, and since $25 is a more common breakfast total, that’s the one that gets the biggest reward.

Or, maybe they just goofed up the math.

Sorry, I know no jokes about gift cards. But here are a couple about finance.

Why didn’t the mathematician report his stolen credit card?

The thief was spending less then his wife.We didn’t exceed the budget. The allocation simply fell short of expenses.

I’m flat broke, so my financial advisor recommended plastic surgery: cut up all my credit cards.

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(Disclosure: I’ve liked A.D. since he was one-and-done at Kentucky. But it wasn’t until last night that I bought an Anthony Davis jersey:

But as much as I like A.D., I couldn’t help thinking that there needs to be an asterisk next to this new All-Star scoring record. If you only look at points, sure, 52 > 42, so Davis scored more points last night than Wilt Chamberlain scored in the 1962 All-Star Game. But that’s only a part of the mathematical story.

First, let’s talk **scoring percentage**. In 1962, the final score of the game was 150‑130, meaning that Chamberlain accounted for 15.0% of all scoring. The final score of last night’s game was 192‑182, meaning that Davis accounted for 13.9% of all scoring. Chamberlain gets the nod, but only slightly, and I’ll admit it’s not insignificant that Davis only played 31 minutes last night, while Chamberlain played 37 minutes in 1962. So, maybe this is a push.

But let’s consider **shooting percentage**. Last night, both teams combined for 55.5% shooting, whereas in 1962, they managed just 43.8% shooting. Perhaps the all-stars from 50 years ago just didn’t shoot as well as players today? Actually, that’s somewhat true: The league FG% for 1961‑62 was 42.6%, the league FG% for 2016‑17 (so far) is 45.6%. But the all-stars last night were 9.9% above the league average, whereas the all-stars in 1962 were just 1.2% above their league average, suggesting that the defense in New Orleans was negligible at best. Which brings me to my next point…

Let’s talk **defense**. Maybe the combined 374 points that were scored last night doesn’t convince you that defense was nonexistent. Then how about this: In the 1962 game, there were 62 personal fouls. Last night, there were only 16. Even more stark, though: In 1962, all-stars shot 95 free throws during the game; last night, they only shot 8. That’s not a typo, and it’s a pretty clear indication that no one was making much effort to contest shots.

Davis played a great game, but it doesn’t feel right that he unseats Chamberlain, given the circumstances. Not to mention, Chamberlain played a more complete game — shooting 73% from the field, grabbing 24 rebounds, and adding 1 assist.

This brings me to my final point, **proportions**. The teams last night scored 1/3 more points than their 1962 counterparts, and if you take away that extra third from Davis, he’d have ended the night with 39 points. So if an asterisk is good enough for Maris’s 61 and Flo-Jo’s 10.49, then it ought to be just fine for Davis’s 52, too.

But it is what it is. Congratulations, Anthony Davis.

Looking at the math of basketball is something I get to do quite a bit these days. Discovery Education has formed a partnership with the NBA, and we’re creating a collection of “problems worth solving” using NBA stats and highlight videos. Wanna see some of what we’ve done? Check out **www.discoveryeducation.com/NBAMath**.

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“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!

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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.

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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.

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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.

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