## Archive for May, 2011

### Ear Among Arms

If you were to eat a starchy tuber this month, it might be said that you are consuming a **May yam**. And if you were to invoke the rule log_{b}(*xy*) = log_{b} *x* + log_{b} *y*, one might say that you are using a **logarithm algorithm**. Both of these phrases are cool, I think, because the second word is an anagram of the first.

The following definitions describe two-word phrases in which the second word is an anagram of the first. The value in parentheses indicates the number of letters in each word.

The first four involve math terms.

- political policy regarding the center of mass (8)
- a clandestine mosquito (6)
- a trendy, golden ratio (3)
- walks around a doughnut (5)

The next two are related to school but are not necessarily mathematical. (Coincidentally, they both happen to be about assessment.)

- educator who supplies students with answers to standardized tests (7)
- high school students who are anxious about the SAT (5)

And these last few are not related to school or math. They’re just for fun.

- jargon associated with website sign-on (5)
- saber, rapier, epee, blade, etc. (5)
- update from a bellhop (6)
- commentary of one who watched (6)
- dormant skill (6)
- when Bambi gets a new app (8)
- relaxed bivalve mollusk (4)
- committee member who checked the Presidential ballots in Florida again (7)
- tedium in the boudoir (7)
- oversized satchel (5)
- stalker who’s already attached (7)

**Answers**

- Centroid Doctrine
- Covert Vector
- Hip Phi
- Torus Tours
- Cheater Teacher
- Tense Teens
- Login Lingo
- Sword Words
- Porter Report
- Viewer Review
- Latent Talent
- Woodland Download
- Calm Clam
- Recount Counter
- Bedroom Boredom
- Super Purse
- Married Admirer

**Bonus question:** The title of this post is also an anagram. Can you decode it?

### Funniest Joke in the World?

I once considered writing a book on how to tell a joke. Primarily, I wanted to write the book for my mother, who knew more jokes than anyone I’ve ever met, but whose delivery left a little to be desired. Ultimately, I decided against it, realizing there might be even less of a market for that book than there is for a collection of math jokes. In addition, I also thought my mother might be offended if I gave her a book on how to tell jokes and said, “Here, I think you should read this.”

Had I written the book, however, one tenet that I would have included is, **make the joke your own.** For instance, a great lawyer joke might be modified to poke fun of statisticians, in which case it might be more appropriate for the next math department social event.

Here’s a simple example of a modification. The following one-liner has recently gained notoriety:

Going to church doesn’t make you a Christian any more than standing in a garage makes you a mechanic.

But perhaps the following modification would fare better with a bunch of professors who are opposed to some states’ overly simplistic requirements for alternative certification of math teachers:

Going to school doesn’t qualify you to be a teacher any more than standing in a garage qualifies you to be a mechanic.

Of course, not every joke can be modified. Some jokes absolutely require a rabbi and a priest, and they wouldn’t be funny with a tortoise and a hare.

With my theories on stealing jokes, modifying them, and passing them off as your own in plain view, allow me to share two of my favorite jokes.

When Richard Wiseman attempted to identify the funniest joke in the world, a slightly different version of the following joke finished second in the voting. For quite some time, it actually held the lead.

A math professor and a graduate student are on a camping trip. After dinner, they retire to their tent for the night. Several hours later, the professor wakes up and nudges his student. He says to the student, “Look up at the sky, and tell me what you see.”

The student replies, “I see millions and millions of stars, sir.”

“And what do you deduce from that?”

The student thinks for a minute. “Well, astronomically, it tells me that there are millions of galaxies and potentially billions of planets. Astrologically, I observe that Saturn is in Leo. Horologically, I deduce that the time is approximately a quarter past three. Meteorologically, I suspect that we will have a beautiful day tomorrow. Theologically, I can see that God is all powerful, and that we are a small and insignificant part of the universe. Why, sir? What does it tell you?”

“You idiot!” the professor exclaims. “Someone has stolen our tent!”

The following is a modification of the world’s funniet joke, which Wiseman found to have universal appeal — it was judged equally fun by men and women, both young and old, from many different countries.

Late one night, two graduate students are working on a problem set. With the solutions due in the morning, both students are stressed. Suddenly, one of them grabs his chest and falls to the floor. He isn’t breathing and his eyes are glazed. The other student quickly dials 911. “I think my friend is dead!” he tells the operator. “What can I do?”

The operator says, “Calm down. I can help. First, are you sure that he’s dead?”

“Just a second,” he says, and puts down the phone. There is a silence, then a shot is heard. Back on the phone, the student says, “Okay, now what?”

### 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.

### The Anagram Game

I have a question.

Is AEGNNTT an anagram of TANGENT?

According to wordsmith.org, it’s not. That site defines an *anagram* as “a word or phrase formed by rearranging the letters of another word or phrase.”

But the game *Anagrams* involves forming words from a random collection of letters.

In the book Word Freak, author Stefan Fatsis calls AEGNNTT an *alphagram* of TANGENT. It’s a rearrangement of the letters in alphabetical order.

Recently, my sons and I have been playing The Anagram Game. Generally, I give them a random collection of letters, so perhaps the name is a misnomer. In any case, they have fun trying to rearrange the letters into real words. Sometimes, I present them with a collection of letters that happens to form a word; for instance, tonight I gave them N-I-G-H-T. They quickly rearranged the letters to form THING. Sometimes, I put the letters into alphabetical order, like A-A-B-B-E-L-L-S. Eli impressed me by decoding that one in less than a second: BASEBALL. But most times, I give them a random arrangement of letters, like U-N-A-T, which can form TUNA or AUNT.

Anyone know what you call an random arrangement of letters that can be rearranged to form a word or phrase? If no such word exists, then I would like to suggest *scramblagram*.

While playing the Anagram Game yesterday, I presented Alex with G-O-N-L. “No, daddy,” he said, “the best anagram of LONG is N-G-L-O.”

Well. I guess I stand corrected.

I asked him, “Is there a best anagram of LAWN?”

He immediately answered, “W-N-L-A.”

“And is there a best anagram of WALK?”

He again answered quickly: “L-K-W-A.”

Interestingly, Eli agreed with him in every case. Apparently my four-year-old sons are privy to a universal truth regarding the reordering of letters in a word, about which I am helplessly unaware.

Luckily, Alex was willing to reveal the pattern. “I always put the two back letters first, then the first letter, and then the vowel.”

So there you have it. The system only works for words of the form consonant-vowel-consonant-consonant, but that’ll do just fine if you’re looking for scramblagrams of LIST, PARK, SULK, BOWL, or RENT.

As it turns out, the word *anagrams* itself has an anagram: *ars magna* (Latin, “great art”).

The following are several of my favorite mathy anagrams.

ELEVEN + TWO = TWELVE + ONE

decimal point = I’m a dot in place

schoolmaster = the classroom

What is the square root of nine? = THREE, for an equation shows it!

### The Humor and Poetry of Jims Maher

Yesterday, I received an email from Jims Maher containing the following joke, which he said he thought up yesterday in a real “facepalm” moment:

There used to be seven bridges in Königsberg.

Two were lost to war. Another two were demolished in peace.

So what does that leave us with?A slippery slope.

Coincidentally, Jims was also the only entrant in the MJ4MF Humorous Math Poem Contest. (I will assume that everyone chose not to submit an entry because I announced the contest on April 1, so all of you thought the contest was a joke. Please allow me to harbor this delusion — it’s easier on my ego that way.) Consequently, a signed copy of *MJ4MF* is on its way to Jims. He said that he plans to “put it to good use as a prize in some fundraiser.” I like your style, Jims!

Because enquiring minds want to know, here is Jims’ award-winning poem…

Start at OneNumbers are counted.

One, two, three…

But some numbers are skipped,

It’s plain to see.We never count zero

Because it’s not there.

And the imaginary numbers

Are as visible as air.It is only the natural numbers

That we will count,

From one on up

To any amount.However, the last number

Can never be known,

Because you can always add one,

However high that you go.And so we keep counting,

From one, to two, to three…

With the natural numbers we keep counting,

From one to infinity.

Forgive the commentary, but I could not help thinking about mathematical definitions when reading Jims’ poem. According to Wolfram MathWorld,

The term “natural number” refers either to a member of the set of positive integers 1, 2, 3, …, or to the set of nonnegative integers 0, 1, 2, 3, …. Regrettably, there seems to be no general agreement about whether to include 0 in the set of natural numbers.

Similarly, the James and James *Mathematical Dictionary* gives three different definitions for *whole numbers*: The set of positive integers 1, 2, 3, …; the set of nonnegative integers 0, 1, 2, 3, …; and the set of all positive and negative integers …, -2, -1, 0, 1, 2, ….

### 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%.

### Pregnancy Probabilities

My twins sons were born on 05/02/07, which is cool because 5 + 2 = 7. Today, they turn 4 years old. Happy birthday, Eli and Alex!

I usually show a picture of them during presentations, and the audience will often ask, “Why are you showing a picture of your sons during a math talk?” Because they’re twins, the response is automatic — “Today we’ll be discussing *multiples*.”

My wife is pushing 40, and I pushed 40 some time ago, and we do not intend to have any more children. Which reminds me of some sage advice:

Women should never have children after 35. I mean, really… 35 kids is plenty!

Seriously, the traditional age at which a woman is considered to be “high risk” is 35. But the following statement from WebMD, like so many others I’ve read, implies that complications are possible when a woman reaches her 35th birthday that weren’t as likely when she was 34 years, 364 days old:

…complications during pregnancy are more common when women reach age 35.

The truth is, the risk of complications during pregnancy increases with the birth mother’s age, so it would also be true to say that complications during pregnancy are more common when women reach age 26, or 33, or 29, or any other age within the child-bearing years.

My wife and I were lucky. We have friends who, over the course of many years, tried multiple fertility treatments without success. By comparison, we were able to conceive in about 10 months, and we didn’t need the help of modern science. But because success was not immediate, Nadine grew worried. One day, she was particularly sad about it, and I tried to cheer her up. “Don’t worry,” I said. “It’ll happen.”

“But I’m *almost* 35!” she screamed.

“Yes?” I said, not quite understanding her point.

“What do you mean, *yes*?” she shouted. “The risk of having a kid with Down syndrome is significantly higher for women over 35.”

I was a little cross to be getting scolded. “That’s true,” I agreed, “compared to a woman who’s 20. But not compared to a woman who’s 34 years, 9 months,” which was Nadine’s age. “If you wanted to significantly reduce the chance of complications, you should have met me a decade earlier.” Needless to say, that comment did not garner a Husband of the Year nomination.

I spent the next several hours collecting data, and later that day, I presented Nadine with the following graph:

Not surprisingly, the likelihood of Down syndrome increases exponentially with the birth mother’s age, and the risk of Down syndrome when the mother is in her mid‑30’s is nearly three times the risk of a mother who is 10 years younger. For a birth mother at age 34, approximately 1 in 617 children are born with Down syndrome; at age 35, approximately 1 in 509 children are born with Down syndrome. So perhaps my statement wasn’t completely accurate. The likelihood of Down syndrome increases nearly 20% when the mother’s age increases from 34 to 35 years.

Nadine and I spent a long time discussing the statistics. I think she felt better when she realized that, even for a mother who is 35, the probability of having a kid with Down syndrome is less than 0.2%.

Speaking of births and twins, the following is a question I’ve posed to various audiences:

In recorded history, what is the greatest number of children born to one woman?

Before I give you the answer, are you sitting down?

Incredibly, the greatest officially recorded number of children born to one woman is 69. She was the first wife* of Feodor Vassilyev (1707‑1782) of Shuya, Russia. (Aren’t you curious why he needed more than one wife?) Unbelievably, 67 of them survived infancy. That would be an amazing mortality rate even today. It’s absolutely astounding for Russia in the middle of the 18th century. What is also unbelievable is that she never gave birth to just one child. Each pregnancy resulted in twins, triplets, or quadruplets. This last fact leads to another question I’ve often posed to audiences:

How many sets of twins, triplets, and quadruplets did she have?

Think about that for a moment, and see if you can find an answer. What you’ll probably realize quickly is that the problem does not have enough information to be solved. So here’s an additional piece of information:

Between 1725 and 1765, she had a total of 27 pregnancies.

Is that enough info to solve the problem?

Actually, it’s not. Even with this additional info, there are still eight possibilities: (19, 1, 7), (18, 3, 6),(17, 5, 5), (16, 7, 4), (15, 9, 3), (14, 11, 2), (13, 13, 1), and (19, 15, 0). So here’s one last piece of data:

She had four times as many sets of twins as quadruplets.

That should do it, and you likely now realize that she gave birth to 16 sets of twins, 7 sets of triplets, and 4 sets of quadruplets. Wow!

* I apologize for not supplying her name. That is not misogyny. Every reference identifies her in relation to her husband but does not give her name.