## Posts tagged ‘MathCounts’

### Problems at the 2016 MathCounts National Competition

Yesterday, Edward Wan (WA) became the 2016 MathCounts National Champion. He defeated Luke Robitaille (TX) in the finals of the Countdown Round, 4-3. In the Countdown Round, questions are presented one at a time, and the first student to answer four correctly claims the title.

“This one is officially a nail-biter,” declared Lou DiGioia, MathCounts Executive Director and the moderator of the Countdown Round. Three times, Wan took a one-question lead; and three times, Robitaille tied the score on the following question. The tie was broken for good when Wan answered the following question:

What is the remainder when 999,999,999 is divided by 32?

This year’s winning question was relatively easy. What makes me say that? Well, for starters, when an odd number is divided by an even number, the remainder will be odd; and because 32 is the divisor, the remainder has to be less than 32. Consequently, the remainder is in the set {1, 3, 5, …, 31}, so there are only 16 possible answers.

But more importantly, most MathCounts competitors will be well trained for a problem of this type. It relies on divisibility rules that they should know, and it requires minimal insight to arrive at the correct answer.

I suspect that the following explanation of the solution is the likely thought process that Wan used to solve this problem; of course, all of this occurred in his head in less than 7 seconds, which does make it rather impressive.

A fact that you probably know:

• A number is divisible by 2 if it’s even.

But said another way…

• A number is divisible by 2 if the last digit is divisible by 2.

There are then corollary rules for larger powers of 2:

• A number is divisible by 4 if the last two digits are divisible by 4.
• For example, we can conclude that 176,432,928 is divisible by 4 because the last two digits form 28, which is divisible by 4. The digits in the hundreds, thousands, and higher place values are somewhat irrelevant, because they represent some multiple of 100 — for instance, the 7 in the ten millions place represents 70,000,000, which is 700,000 × 100 — and every multiple of 100 is divisible by 4.
• A number is divisible by 8 if the last three digits are divisible by 8.
• For example, we can conclude that 176,432,376 is divisible by 8 because the last three digits form 376, which is divisible by 8 since 8 × 47 = 376.
• A number is divisible by 16 if the last four digits are divisible by 16.
• A number is divisible by 32 if the last five digits are divisible by 32.
• And so on.

These observations lead to a generalization…

• A number is divisible by 2n if the last n digits are divisible by 2n.

I won’t take the time to prove that statement here, but you can trust me. (Or maybe you’d like to prove it on your own.) I will, however, explain why it’s relevant.

A number will be divisible by 32 if the last five digits are divisible by 32. Consequently, any number that ends in five 0’s will be divisible by 32, which means that 1,000,000,000 is a multiple of 32. Since 999,999,999 is 1 less than 1,000,000,000, then it must be 1 less than a multiple of 32. Therefore, when 999,999,999 is divided by 32, the remainder will be 31.

The hardest part of solving that problem is recognizing that 999,999,999 is 1 less than a multiple of 32. But for most MathCounts students, that step is not very difficult, hence my contention that this was a relatively easy winning problem.

My favorite problem of the Countdown Round? Now, that’s another story, and it epitomizes what I generally love about MathCounts problems.

If a, b, c, and d are four distinct positive integers such that ab = cd, what is the least possible value of a + b + c + d?

This problem has several things going for it:

• It’s simply stated.
• It’s easily understood, even by students who don’t participate in MathCounts.
• It has an entry point for all students, since most kids can find at least one set of numbers that would work, even if they couldn’t find the set with the least possible sum.
• Finding the right answer requires convincing yourself that no lesser sum exists.

It’s that last point that I find so interesting. While I was able to find the correct answer, it took a while to convince myself that it was the least possible sum. But since I don’t want to deprive you of any fun, I’ll let you solve the problem on your own.

As a final point, I’ll show you a picture that I took at the event. Do you see the error? What can I say… it’s a math competition… you didn’t expect them to be good with numbers, did you?

Full disclosure: The error was corrected halfway through the competition during a break.

### Making Progress, Arithmetically

Today is 11/12/13, a rather pleasant-sounding date because the numbers form an arithmetic sequence, albeit a trivial one. It’s not the only date in 2013 for which the month, date, and year form an arithmetic sequence. How many others are there?

Several nights ago, my sons asked if they could do bedtime math, but Eli asked if we could do problems other than those on the Bedtime Math website, because “they’re a little too easy.” So instead, I navigated to the MathCounts website and opened the 2013-14 MathCounts School Handbook. We scrolled to page 9 and attacked the problems in Warm-Up 1.

Things were going well until we reached Problem 8 in the set, which read:

The angles of a triangle form an arithmetic progression, and the smallest angle is 42°. What is the degree measure of the largest angle of the triangle?

I could have just answered Eli’s question by stating the definition:

An arithmetic progression is a sequence of numbers for which there is a common difference between terms.

But such a definition isn’t very helpful, since I’m not sure that either Eli or Alex know what sequence, common difference, or term mean. It would have led to even more questions.

Plus, I’ve always believed that kids understand (and retain) more when they discover things on their own. Call it “discovery learning” or “inquiry-based instruction” or any of myriad other names from educational jargon, it just means that giving kids the answer is not the most effective way for them to learn.

So instead, I said, “Let me give you some examples.” And then I wrote:

1, 2, 3

3, 5, 7

Alex said, “Oh, I get it! An arithmetic progression is a nice pattern of numbers.”

So I said, “Well, let me give you some patterns that aren’t arithmetic progressions.” And then I wrote:

2, 4, 8

“That’s a nice pattern, isn’t it?” I asked. “But it’s not an arithmetic progression.”

“Oh,” said Alex. He thought for a second, then revised. “You have to add the same amount every time.”

And there you have it. Three examples, and my sons were able to define arithmetic progression. It’s not as sophisticated as “a common difference between terms,” but “add the same amount every time” is a sufficient definition for a six-year-old.

So they generated an arithmetic progression with 42 as the smallest term:

42, 45, 48

Eli said, “I don’t fink vat’s enough.” When asked to explain, he said he thought that the angles in a triangle add up to 180 degrees.

“Are you sure?” I asked. He wasn’t. Nor was Alex. So I asked if they could convince themselves that the sum of the angles is 180°.

Alex said, “Well, the angles in a square add up to 360°, and you could cut it in half.” So we did:

They then reasoned that each triangle would have a sum of 180°. “But maybe that only works for a square,” I said. “How do you know it’ll work for other shapes?”

Eli suggested that we could cut a rectangle in half, too:

And again they concluded that each triangle would have a sum of 180°.

Understand, this is NOT a proof of the triangle sum formula. When they get to high school and need to demonstrate the rigor that the Common Core State Standards are demanding, well, then we’ll worry about formal proof. But for now, I’m okay with six-year-olds who can demonstrate that kind of reasoning.

They then took another guess, but this time they chose three numbers that added to 180:

42, 59, 79

Realizing that the difference between the first and second terms was 17 and the difference between the second and third terms was 20, they revised:

42, 60, 78

They concluded that the largest angle had a measure of 78°. And all was right with the world.

So why am I telling you all this?

Partially, it’s because I’m a proud father.

But more importantly, it’s because this vignette demonstrates that teaching is an art, and successful teaching doesn’t happen by accident. It’s not easy, as many people believe. What’s easy is the perpetuation of bad teaching, a la Charlie Brown’s teacher, or textbooks that simply present information with the belief that students will absorb it by osmosis. Good teaching, however, requires content knowledge and pedagogical knowledge, and it demands teachers who can handle unexpected classroom twists and turns and have the ability to adjust on the fly.

A student is convinced that a right triangle isn’t a right triangle because the right angle isn’t in the lower left corner? You better find an effective way to clarify that misconception. (Hint: Don’t use a traditional textbook where every picture of a right triangle shows the right angle in the lower left corner.)

Students think that 16/64 = 1/4 because you can “cancel the 6’s”? Uh-oh. Better find some counterexamples pronto, and help them understand why 16/64 can be reduced to 1/4.

Your students don’t know the definition of arithmetic progression? Then you better figure out a way to help them define it, and just writing your definition on the chalkboard isn’t gonna cut it.

Want to see what good teaching looks like? See Dan Meyer, or Christopher Danielson, or Fawn Nguyen. Or many, many others who don’t blog about it but inspire students every day.

Someday soon, I hope to add my project at Discovery Education to the list of examples of good teaching. Until then, I’ll just keep blathering about my sons.

### Fun and Sun at 2013 MathCounts National Competition

On May 10, I attended the MathCounts National Competition in Washington, DC.

From my perspective, the most important part of the event was the world-record attempt to place 244 middle school students, 61 coaches, and 20 advisors into an arrangement of the first 25 rows of Pascal’s Triangle. (As the chair of the MathCounts Question Writing Committee, I was counted among the advisors.) This may sound easy — after all, each term of Pascal’s Triangle is the sum of the two numbers above it, and the triangle is vertically symmetric. How hard can it be to arrange people holding the numbers?

Well, there were two pieces to this attempt that increased the difficulty level. First, no one knew what number they would get ahead of time. Each of us received a sealed manila envelope, and inside was a number from Pascal’s Triangle printed on a piece of 8.5″ × 11″ cardstock. We weren’t allowed to open our envelopes until given the signal to begin.

Second, the folks from Guinness required that we accomplish this feat in under 10 minutes. (Unbeknownst to me, if you want to set a benchmark for a Guinness World Record, you can’t just call the folks at Guinness and tell them that you were able to remove all the hairs from your chihuahua using only a pair of tweezers in 3 days, 16 hours, and 34 minutes, for example, and expect that they’ll include you in the book. Instead, you have to call Guinness ahead of time, tell them what you’re planning to do, and then they’ll tell you how fast you’ll have to do it for them to consider it worthy of inclusion.)

An adjudicator from Guinness was there to officiate and ensure that there was no funny business. He confirmed that all 25 rows were assembled correctly, and because we were able to complete the triangle in an astounding 6 minutes, 16.5 seconds, we set a Guinness World Record. This is what we looked like upon completion:

I’m along the right side of the triangle, holding a 9. You may not know what I look like, but I’m still fairly easy to spot — even sitting down, I’m quite a bit taller than the middle school students who surrounded me.

[Update 5/21/13: A video of the attempt was uploaded to YouTube by one of the participant’s father. Major kudos to the MathCounts staff and the folks at the Art of Problem Solving, who devised the clever scheme used to shuffle folks into their appropriate places.]

The Countdown Round — a Jeopardy-like event where twelve kids, arranged in a tournament bracket format, compete head-to-head to answer questions faster than their opponent — is the culmination of the event. The winning question from this year’s Countdown Round was:

What is the greatest integer that must be a factor of the sum of any four consecutive positive odd integers?

Alec Sun was crowned national champion when he gave the correct answer to this question. (Which he did in less than 20 seconds, by the way. You can have more time than that, but you won’t find the answer on this page.)

The Countdown Round was moderated by Lou DiGioia, executive director of MathCounts. During the round, he set up a fantastic joke.

When Akshaj Kadaveru, a seventh-grader from Virginia, took the stage for the Countdown Round, Lou asked him, “You’re a local. What’s one place in Washington, DC, that you’d recommend your fellow competitors see during their visit to the nation’s capital?”

Akshaj suggested the Smithsonian Air and Space Museum.

“Oh, I’m sorry,” said Lou. “The answer we were looking for is the zoo.” And he held up a piece of paper with the word ZOO.

A little while later, another competitor from Virginia, seventh-grader Franklyn Wang, took the stage. Lou asked him the same question about which place out-of-town competitors should visit.

“This hotel!” Franklyn giggled into his microphone.

“Actually,” said Lou, “the correct answer is the zoo.” Again, Lou held up a piece of paper with the word ZOO.

When Alec Sun took the stage, Lou explained to the crowd that Alec was the only competitor in the history of MathCounts to make it to the Countdown Round three straight years. “But,” Lou said to the audience, “I was talking to Alec yesterday, and he told me that he was nervous. In his two previous appearances, he hadn’t answered a single question. I’d really hate for him to leave the stage today without having answered at least one question correctly. So, Alec,” he said, turning to the young man, “I’m going to give you a non-math question to answer. It’s not a softball — you’re really gonna have to think about this one. But I’m confident you can do it.” And then Lou asked him, “Alec, what is one place you’d recommend that folks visit while they’re in town this weekend?”

“Correct!” yelled Lou, holding the piece of paper in the air. “The zoo!” And the entire room exploded into applause.

Who said that math people can’t be funny?

### Hardest Question at MathCounts 2012?

Rumor has that some big professional football competition will occur on Sunday, but I’ll be setting my sights on a more intellectual competition that takes place on Saturday. Middle school students across the country will be participating in the MathCounts Chapter Competition. Good luck to all mathletes!

As a member of the MathCounts Question Writing Committee, one of my responsibilities is to ensure that the questions on the competition are not too hard, not too easy, but just right. Consequently, the committee reviews a lot of data regarding the percentage of students that answer each question correctly.

Rather accidentally, I wrote the most difficult question that appeared on the 2012 Chapter Competition.

The chart below shows the number of students from a random sample (n = 1200) who answered each of the 30 questions correctly on the Sprint Round:

Although Problem 29 was answered correctly by just 42 students, it was not the most difficult. Students are given 40 minutes to complete all 30 problems in the Sprint Round. Consequently, many students never get to the problems at the end of the round, and the last several questions are typically answered correctly by only a handful of students.

On the other hand, Problem 14 was answered correctly by only 50 students, and it occurs in the middle of the round. By design, the questions in the round proceed from least difficult to most difficult, so problem 14 was intended to be a medium question. Questions on either side of Problem 14 were answered correctly by 500 to 700 students, and the members of the question writing committee suspected that Problem 14 would be answered correctly by a similar number. Boy, did we miss the mark!

(You’ll notice that we also incorrectly guaged the difficulty of Problem 9, though not quite as dramatically.)

So, what was this doozie of a question? Try it for yourself:

If Friday the 13th occurs in a month, then the sum of the calendar dates for the Fridays in that month is 6 + 13 + 20 + 27 = 66. What is the largest sum of calendar dates for seven consecutive Fridays in any given year?

What do you think? Are you as smart as the 4.167% of mathletes who got this problem correct? To find out, take a look at the calendar for May and June 2013.

### 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 n such that a square cannot be partitioned into n smaller 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%.

### Inversions

I recently heard a first-grade teacher espouse the benefits of having students create equations that are both true and false. Her argument was that students typically only see equations that are true, but by considering equations that are false, students develop a better understanding of equality and the equals sign.

She showed us a stack of equations, both true and untrue, that her students had created. One of them read:

6 – 6 = 9 – 9

This equation happens to be true, but what struck me was its symmetry. I commented, “That one is especially nice, since it’s also true when you display it upside down.” Of course, this is a trivial example of an equation that remains true when rotated 180 °, but it led me to consider a bigger challenge:

Can you find a non-trivial example of an equation that remains true when rotated 180°?

And…

What is the most complex example of an equation that remains true when rotated 180°?

I think today (9/6) is a good day to share this puzzle with you, since today’s date, minus the year, remains the same when rotated.

I was able to discover a few (fairly basic) answers to the first question, but I won’t share them here for fear of spoiling your fun. As for the second question, I don’t have an answer; in fact, I’m not even sure it’s answerable.

Here’s a poem to remember the procedure for dividing fractions, which happens to reference inversion:

Ours is not to reason why;
Just invert and multiply!

Scott Kim is the undisputed master of the art form known as Inversions; if you haven’t seem them before, you should definitely check out some of his inversions.

An inversion is a word or phrase that reads in more than one way. Most often, an inversion involves writing letters so that they have rotational or reflexive symmetry — that is, the word or phrase reads the same when turned upside down or when viewed in a mirror.

I’m a rank amateur, and my best inversion is shown below.

Scott Kim created an inversion of my name, too, and his version is shown below.

I think mine is better, but there are several major difference between the two inversions:

• I created mine on a computer, so it has perfect rotational symmetry and nice clean lines. Scott Kim drew his freehand.
• My version is only my first name. Scott’s version contains both my first and last names.
• The inversion that I created took several weeks and many refinements to complete. The inversion created by Scott Kim took him less than 30 seconds. (No, seriously — he did it on the spot after I spelled my last name for him.)

Finally, let me tell you the story about the first time I met Scott Kim. I used to work for the MATHCOUNTS Foundation, and one of my responsibilities was finding a speaker who could entertain 228 middle school students each year at our national competition. In 1996, I contacted Scott, and he agreed to give a talk. I was ecstatic that I was able to convince him to speak to our participants, but at our weekly staff meetings leading up to the national competition, my boss would continually question my choice. “Who is he?” Camy would ask. “What does he do?” At five consecutive staff meetings, I tried to explain what it is that Scott Kim does. I even showed her examples, but she just never got it. Finally, at the sixth staff meeting, the questions came again, and I finally responded, “I’m sorry that I’m not able to get you to understand how cool he is. But you’ve got to trust me, Camy — the kids are going to love him.”

When Scott Kim arrived at the national competition, I was unfortunately busy, so I quickly introduced him to my boss, and then I excused myself. Consequently, I wasn’t present for the rest of the story, but this is how it was relayed to me.

“Camy,” said Scott Kim. “That’s an interesting name. I don’t think I’ve heard it before. Is that with one M or two?”

“Just one,” Camy said.

Scott thought for a moment, and then scribbled something on his notepad. He ripped off the sheet and handed it to her.

She looked at it. “Yes,” she said. “C-A-M-Y.” She was clearly unaware of what she was looking at.

Scott said, “Actually, it’s not just your name. It’s an inversion.” He then turned the paper upside down for her, and there was her name again.

“Oh, my God!” she screamed. “That is so cool!”

Ha-rumph. In less than five seconds, Scott Kim was able to do what I couldn’t do in five consecutive staff meetings.

But it gets better. After chatting for a few minutes, Camy introduced Scott Kim to another celebrity who was attending the competition — Scott Flansburg, who is better known as The Human Calculator. Scott Flansburg had just witnessed what Scott Kim had done with Camy’s name, so when Scott Kim scribbled something on a piece of paper and handed it to him, he looked at it and said, “Oh, I get it. It says Scott Flansburg, and when I turn it upside down, it’ll say Scott Flansburg again.”

“No,” said Scott Kim. He then turned the paper upside down for Scott Flansburg, who read aloud what was now showing on the paper: “The Human Calculator.”

That’s right. In less than 20 seconds, Scott Kim created an inversion that read “Scott Flansburg” in one direction and read “The Human Calculator” when rotated 180°.

Wow. Unbelievable.

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

## MJ4MF (offline version)

Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.