## Posts tagged ‘squares’

### Book Review: The Joy of *x*

Steven Strogatz says that by arranging things in the right way, we can make a surprising link seem obvious — the hallmark of an elegant proof.

In particular, he’s talking about arranging a group of rocks — several groups of rocks, actually, each containing an odd number — to show that when you add all of the consecutive odd numbers, starting with 1, the sum is a square number:

1 + 3 = 4

1 + 3 + 5 = 9

1 + 3 + 5 + 7 = 16

1 + 3 + 5 + 7 + 9 = 25

I didn’t have to read *The Joy of x* to know that 1 + 3 + 5 + 7 + … + (2*n* + 1) = *n*^{2}, and of course I’ve seen the visual proof that Strogatz demonstrated with a group of rocks:

But I don’t think that I had ever heard a definition for the concept of an elegant proof. Therein lies the beauty of Strogatz’s new book: while most of the concepts he covers will be familiar territory for the mathy folks who read this blog, each of the 30 chapters contains a pearl of wisdom, or an interesting factoid, or an eloquent sentence that makes you realize he’s as good with words as he is with numbers.

For instance, you probably didn’t know this about the distribution of heights on the online dating site OkCupid:

…the heights reported by both sexes follow bell curves, as expected. What’s surprising, however, is that both distributions are shifted about two inches to the right of where they should be. So either the people who join OkCupid are unusually tall, or they exaggerate their heights by a couple of inches when describing themselves online.

The chapters within the book are short and, as Strogatz admits, are not dependent on one another. You could start at Chapter 10, “Working Your Quads,” and it wouldn’t matter if you had read any of the nine chapters that precede it.

All in all, *The Joy of x* is breezy and fun, and while it won’t significantly further the education of a mathematician, it will provide some entertainment. And for the non-mathematician, it provides a nice overview of mathematics “from one to infinity,” covering everything from numbers to geometry to calculus, giving adults a second chance at the subject.

### Super Bowl Math (and Results of the Super Bowl Squares Online Contest)

There were a lot of interesting mathematical things that happened tonight.

First things first: Big props to Valerie Strauss of The Answer Sheet, who asked, “Tom Brady vs. Eli Manning: Who’s Smarter?” and then was smart enough to link to one of my previous posts when trying to answer the question.

Math Incident #1

Within the first five minutes of coverage, it was announced that the NFC had won 14 consecutive coin tosses. Posted on the screen:

Odds: 1 in 16,384

I was a little bummed that it didn’t say, “Odds: 1 in 2^{14}.” But I can’t complain. It’s not every day that probability gets international publicity.

Math Incident #2

With just under 4:00 left in the game, Wes Welker dropped a pass from Tom Brady. During the replay, Cris Collinsworth said that it was a pass that Welker makes “100 times out of 100.” Um, Cris, in case you missed it… I don’t know how many passes just like this that Wes Welker has caught, but he missed this one, so we have at least one data point showing that, in fact, he doesn’t always catch this pass. If you want to revise your statement to “99 out of 100,” I could live with that.

Super Bowl Squares Online Contest

The results of the Super Bowl Squares Online Contest have been posted at http://mathjokes4mathyfolks.com/super-bowl-squares-results.html. But allow me to spoil some of your fun before you click that link:

- There were no winners for the first quarter score (Patriots 0, Giants 9; winning square, 9‑0).
- There were no winners for the second quarter score (Patriots 10, Giants 9; winning square, 9‑0).
- There were no winners for the third quarter score (Patriots 17, Giants 15; winning square, 7‑5).
- There were
**two**winners in the fourth quarter (Patriots 17, Giants 21; winning square, 7‑1).

That means that Ben Morris and Tom Coffin were the only winners of the 31 participants, so they split the $155 pool of Monopoly money, each receiving $77.50.

### Super Bowl Squares Contest

Laurence Tynes, the hero; Billy Cundiff, the goat. And so we head to Super Bowl XLVI with a rematch of the game four years ago. One can only hope that this game will be half as exciting as that one.

Your math/football trivia for the day? **Super Bowl XLVI is the second to require each of the first four Roman numerals (I, V, X, L)**; the first was Super Bowl XLIV two years ago. [Thanks to Eric Langen for pointing out my previous error.] Personally, I’m looking forward to Super Bowl LXVI, when the first four Roman numerals will occur in decreasing order. A real treat will occur in 3532, when Super Bowl MDLXVI will be played, wherein all six of the Roman numerals will appear in decreasing order. While I’m fairly certain I won’t be around to see that one, I hold out hope that I am reincarnated as a star football player who earns that game’s MVP honors; though it’s far more likely that I will return as a football to be used by adolescents in a backyard game.

Buoyed by the success of the online version of my favorite game, I’ve decided to run another online contest. This one relates to Super Bowl XLVI, and you’re asked to predict the units digit of each team’s score at the end of each quarter when the Patriots and Giants square off on Sunday, February 5.

Probably the most common type of office betting pool is a square football pool, which is often referred to as just The Squares. The pool is played on a 10 × 10 grid, and contestants can buy squares within the grid for a certain amount of money. **After all 100 squares have been purchased**, the numbers 0‑9 are randomly assigned to each row and column. The numbers for each row represent the units digit of the score for one team, and the numbers for each column represent the units digit of the score for the other team. The winners are the four people whose squares correspond to the units digit of the actual score of the game at the end of the 1st, 2nd, 3rd, and 4th quarters.

Feel free to use this Excel spreadsheet if you’d like to run your own version of this game. (Though be sure to check all applicable laws, to ensure that you’re not in violation of local or state gaming laws.)

The difference between the typical version of this game and the version I’m running here is that **you get to pick which pairs of numbers you want**. Consequently, winning isn’t solely a matter of random luck. But there’s a catch — you can pick the most likely number pairs, but chances are other folks will pick those numbers, too, and the winnings are divided among everyone who picked that pair. So, should you pick 0‑0 and divide the pot with a thousand others; or should you pick the highly unlikely 5‑2 and have the winnings all to yourself?

*Please note that the game I’m running is for entertainment only. No money is required to play, and there will be no pay-out to the winners.* If all goes well this year, perhaps next year there will be a real version that allows you to wager your hard-earned money in such a silly manner — assuming, of course, that I can find a way to skirt the myriad state gaming laws that would prevent me from running such a contest.

In case you’re wondering, “Why are you doing this?” remember that I’m the author of a math joke blog. Why do I do any of the things I do? For fun, mainly, and because I’m a certifed math geek. I like the math psychology of this game, and I’m just interested in the numbers that people will pick.

Here are the official rules:

- Imagine that you have $5, and each square costs $1, so you can buy up to five squares. It’s your money, spend it how you like — if you want to choose the same pair of numbers for all five bets, go ahead, knock yourself out. And what the hell do I care? Enter as often as you like; if you’ve got nothing better to do with your time than repeatedly submit entries for this contest, well, that’s your problem.
- All money bet will be divided equally among the four quarters, so the total amount will be equal to $5
*n*, where*n*is the number of contestants. (Should a contestant enter fewer than five choices, the last entered choice will be repeated multiple times to get the total to five.) - If you pick a winning square, you will share the winnings with everyone else who picked the same square. (For example, if 200 people play this game, there will $1,000 in the pot, so the winning amount for each quarter will be $250. If ten people choose 7-3 and it hits for one quarter, each person will receive $25.)
- Enter your five choices as two-digit numbers,
**where the tens digit represents the Patriots’ score and the units digit represents the Giants’ score**. (For instance, if you want Patriots 7, Giants 3, enter 73; but if you want Patriots 0, Giants 7, enter 07.)

That’s it. Access the form via the link below:

My friends Andy and Casey Frushour have been keeping data about which pairs of numbers occur most often. Before making your picks, you might want to check out their analysis of data from six years of NFL games as well as from all 45 Super Bowls.

Bets will be accepted until 11:59 p.m. ET on Saturday, February 4, and an image showing the number of times each square was chosen will be posted at:

**Super Bowl Squares Contest – Summary of All Bets**

The complete results for this contest will be posted on Monday, February 6, at the URL below. (But note that this link will return a “404 Error – File not Found” message prior to February 6.)

**Super Bowl Squares Contests – Results**

Good luck!

### Wu, California Earn MathCounts Titles

Today, I had the privilege of attending the MathCounts National Competition. I served as a volunteer in the scoring room, and I had the pleasure of scoring the competitions of the winning team. Congratulations to the California team of Celine Liang, Sean Shi, Andrew He and Alex Hong, who scored an amazing 59.5 (out of 66) on the competition.

You may be thinking that 90.1% correct is not very impressive. With the current level of grade inflation, I suppose that’s a reasonable reaction. But given that these students solve 48 non‑trivial questions (see examples below) in under two hours, it’s an amazing feat. Given twice as much time, most of us would be lucky to answer half of them correctly.

During the Target Round of the competition, students are given six minutes to solve a pair of problems. My favorite problem on this year’s competition was the last question in the Target Round:

How many positive integers less than 2011 cannot be expressed as the difference of the squares of two positive integers?

It’s a good question. You might like to try solving it. But don’t be too discouraged if it takes you more than six minutes… after all, no one expects you to be as smart as a middle school student.

One of the California competitors, Sean Shi, scored well enough in the written competition to qualify for the Countdown Round. The Countdown Round is an NCAA‑style bracket for the top twelve competitors. All students who qualify for the Countdown Round are introduced to the audience, and the following story was told about Sean:

Sean’s younger brother is two years younger than Sean. When Sean was 7 and his brother was 5, his brother told their mom, “You’re the best mommy in the whole world!” Already a mathematical prodigy, Sean replied, “Actually, the chance of that being true is very low.” Sean explained, “No offense, mom! I’m just being mathematical!”

Another competitor in today’s Countdown Round, Shyam Narayanan of Kansas, finished among the top four in last year’s national competition. As a result, he was invited to meet President Obama in an Oval Office ceremony. During the visit, Shyam asked the President a question that he said he had never been asked before.

Since the Oval Office is an ellipse, where are its foci?

Although Obama didn’t know the answer, the mathletes in attendance were able to help him figure it out.

Though Shi and Narayanan made respectable showings in the Countdown Round, it was Scott Wu from Louisiana who prevailed and earned the title of national champion. Wu, who’s older brother Neil Wu was the 2005 MathCounts National Champion, defeated Yang Liu 4‑1 in the finals. Wu locked up the title by correctly solving a rather odd question about digits.

It takes 180 digits to write all of the two-digit positive integers. How many of the digits are odd?

In the Countdown Round, students only have 45 seconds to answer each question — though typically, they answer the questions much faster. So how do you come up with an answer to this question in just a few seconds?

First, realize that exactly half of the two‑digit numbers have an odd units digit. Then, notice that 50 numbers have an odd tens digit, but only 40 have an even tens digit. Consequently, of the 180 digits required, there will be 10 more odd digits than even. Hence, 95 digits are odd, and 85 digits are even.

### MathCounts National Competition

Three days from now, more than 200 mathletes will compete in the MathCounts National Competition in Washington, DC. The 228 mathletes — four each from the 50 U.S. states and various territories — will descend on the Renaissance Washington DC Downtown Hotel and attempt to solve 48 problems in three hours. If you happen to be in the nation’s capital and can spare a few hours, it’s worth attending the event. The students who attend are so talented, mature and well educated, it’s easy to forget that they’re only in middle school. (Coincidentally, the hotel has a wonderfully numerical address: 999 Ninth Street NW, Washington, DC 20001. Four 9’s in the street address and three 0’s in the ZIP code — how awesome!)

As a member of the MathCounts Question Writing Committee, I’ve authored 170 problems and reviewed more than 700 problems in the past eight months. A subset of those questions will appear on the 2012 MathCounts school, chapter, state, and national competitions.

To show their appreciation for our work, the MathCounts Foundation recently sent a shirt to all members of the QWC. Here it is:

On the front, it says:

**I CREATE PROBLEMS**

On the back, it says:

**SEE IF YOU CAN SOLVE THEM**

**MathCounts Question Writing Committee 2010–11**

Among the questions that I reviewed this year are some absolute gems that I’d love to share, but for obvious reasons, I cannot. However, I can share a few of my favorite problems from the book The All-Time Greatest MathCounts Problems, which I edited with Terrel Trotter, Jr.

The following problem appeared on the Sprint Round of the 1992 National Competition. During the Sprint Round, students solve 30 problems in 40 minutes, which is an average of just 1 minute, 20 seconds per problem.

What is the greatest number of bags that can be used to hold 190 marbles if each bag must contain at least one marble, but no two bags may contain the same number of marbles?

Think about it for a second — or 80 seconds, like the competitors do — then read on.

When this problem was originally placed on the 1992 National Competition, the members of the QWC believed the answer to be 19 bags: 1 marble in the first bag, 2 marbles in the second bag, 3 marbles in the third bag, and so on, with 19 marbles in the nineteenth bag (1 + 2 + 3 + … + 19 = 190). At the competition, one student protested, claiming that the answer was 190 bags. On the protest form, he drew a picture like this:

The student explained: Put one marble in the first bag. Then put one marble *and the first bag with its marble* inside the second bag; hence, the second bag now has two marbles. Continue in this manner for 190 bags.

Consequently, the alternate answer of 190 bags was accepted by the judges. This is especially noteworthy, since the problem had been reviewed by nearly 40 people, including university professors, secondary math teachers, and engineers, and none of them had considered this alternate solution.

The next problem is from the 1999 National Competition. It appeared on the Team Round, which means that it was one of 10 problems that four students attempted to solve in 20 minutes.

As shown below, a square can be partitioned into four smaller squares, or nine smaller squares, or even into six or seven smaller squares provided the squares don’t have to be congruent. (Note that overlapping squares are not counted twice.) What is the greatest integer

nsuch that a square cannot be partitioned intonsmaller squares?

Any square can be divided into four smaller squares, thus increasing the total number of squares by three. The first figure shown above is divided into four smaller squares; when the lower left square within that figure is divided into four even smaller squares, the result is the last figure shown above, which is divided into seven squares. By dividing one of the smaller squares of that one into four would yield a figure divided into 10 smaller squares, then 13, 16, 19, …, accounting for all values of *n* ≡ 1 mod 3 for which *n* > 1.

Also shown above is a square divided into six smaller squares. Dividing one of the smaller squares into four would yield a figure divided into 9 smaller squares, then 12, 15, 18, 21, …, accounting for all values of *n* ≡ 0 mod 3 for which *n* > 3. That leaves only the case for *n* ≡ 2 mod 3, for which there are no examples shown above. But producing a square divided into eight smaller squares is easy enough to do:

Dividing one of the smaller squares into four would yield a square divided into 11 smaller squares, then 14, 17, 20, 23, …, accounting for all values of *n* ≡ 2 mod 3 for which *n* > 5.

The only number of squares not on our list are 2, 3, and 5. Consequently, *n* = 5 is the greatest integer such that a square cannot be partitioned into *n* smaller squares.

The final round of a MathCounts competition is the Countdown Round, a *Jeopardy!*‑like event in which two students compete head-to-head to answer questions that are projected on a screen. Students are given just 45 seconds per problem. Because the questions are expected to be answered in a short amount of time, they are not as challenging as those in other rounds. That said, they’re not exactly easy, either! In 1995, the final question of the Countdown Round had a counterintuitive answer, so it garnered a lot of attention:

Out of 200 fish in an aquarium, 99% are guppies. How many guppies must be removed so that the percent of guppies remaining in the aquarium is 98% ?

Originally, there must have been 200 × 0.99 = 198 guppies. If *x* guppies are then removed, the ratio of guppies remaining is (198 – *x*) / (200 – *x*). Since 98% of the remaining fish must be guppies, the following equation results:

(198 – *x*) / (200 – *x*) = 98/100

Solving for *x* yields *x* = 100.

Although the algebra is not difficult, most folks find it counterintuitive that 50% of the fish have to be removed from the aquarium to reduce the percent of guppies by just 1%.