## MathCounts Problems, Practice, and New Friends

Later today, my sons will represent Sellwood Middle School at the Oregon State MathCounts competition; and, if they do well enough, they’ll represent Oregon at the 2021 Raytheon Technologies MathCounts National Competition. They were invited to compete in the state competition today because they finished first and second in the local MathCounts competition:

The local MathCounts organizers hosted a virtual celebration for participants, and at the end, I asked if any students from other schools who qualified for the state competition would be interested in some joint practice sessions. As a result, the coach and two students from Access Academy joined Alex, Eli, and me for two 90-minute sessions this past week, during which we solved some previous state- and national-level problems. One with which we had great fun and lots of discussion was about cryptocodes:

A certain cryptocode must contain one letter from the set {X, K, M, Z} and three distinct letters from the set {W, X, Y, Z}. The four letters can be arranged in any order, and since X and Z are in both sets, these letters may each appear twice in an arrangement. How many cryptocodes are possible?

2017 MathCounts National Competition, Target Round, Problem 8

When asked for their answers, Alex suggested a number that was 24 too low, and Eli gave a number that was 24 too high. After discussing the solution, the other coach said, “Alex and Eli, I think it’s awesome that the average of your answers was the correct answer! Is that because you’re twins?” Now, that’s funny.

A video solution from Sjoberg Math is available on YouTube; my solution is below.

I’m occasionally asked how to prepare for MathCounts competitions. Our in-home preparation program involved several parts:

- Leave math and puzzle books — such as those written by Ben Orlin, Alex Bellos, and Martin Gardner — on the living room table for them to discover.
- Watch YouTube videos — such as those from Numberphile, Matt Parker, and 3Blue1Brown — to see that math can be fun (and that math people can be funny).
- Talk about math and solve problems at the dinner table. One of our favorites is determining the number of clinks that happen after a toast, when everyone at the table offers “Cheers!” and taps glasses with everyone else.
- Use the Art of Problem Solving‘s MathCounts Trainer. Of note, my sons discovered this on their own and started using it because they enjoy solving problems, not because they were training.
- Complete a few MathCounts competitions from previous years. This is, in fact, the only part of the regimen that was actual training, and all students in our math club did two practice competitions prior to the local competition. The main purposes were to expose them to the types of questions on MathCounts competitions; to prepare them for the intensity of the competition (they are presented with 38 questions to be attempted in about 90 minutes); and, most importantly, to prepare them for the reality that they likely won’t get all of the questions correct, which, for most math club students, stands in stark contrast to their performance on the assessments they complete in their regular math class.

I offer this list to anyone who is coaching or interested in coaching a MathCounts team. The purpose of MathCounts is to get students excited about math; the stated mission is “to build confidence and improve attitudes about math and problem solving.” Winning may be fun, but it’s not the goal. To quote Boris Becker, “I love the winning. I can take the losing. But most of all, I love to play.” MathCounts provides students a chance to play with math, and most of them won’t win. Still, it’s an amazing opportunity to show kids how much fun math can be.

—

There are 16 ways to choose the four letters for a cryptocode. The codes in blue text (eight combinations in the middle columns) can each be arranged in 4! = 24 ways.

XWXY XWXZXWYZ XXYZ | KWXY KWXZ KWYZ KXYZ | MWXY MWXZ MWYZ MXYZ | ZWXY ZWXZ ZWYZ ZXYZ |

The codes in green text (six combinations in the first and last column) have a repeating letter (either X or Z), so they can be arranged in 4!/2! = 12 ways each.

Finally, the codes in **red bold text** (one combination in the first and last column) can also be arranged in 4! = 24 ways — but watch out! They’re the same sets, both consisting of W, X, Y, and Z. So only count that set once, not twice.

In total, then, there are 9(24) + 6(12) cryptocodes. Alex explained that this could be computed by rewriting it as 18(12) + 6(12) = 24 × 12, and one of the students from Access Academy rewrote it as 9(24) + 3(24) = 12 × 24. Both obviously reveal the answer, 288 cryptocodes.

I’m 99% certain that that’s the correct answer. And I’m 100% certain that it’s **two gross**.

## Happy as L Solutions

In celebration of my 50th birthday, on Monday I published ten problems involving the number 50. In case you missed, here they are again:

- There are 50 puppies to be adopted at a shelter, and 98% of them are hounds. How many hounds must be adopted so that 90% of the remaining puppies are hounds?
- Let A = 1, B = 2, …, Z = 26. Find two common English words for which the product of the letters is 50.
- What’s the least possible product of two prime numbers with a sum of 50?
- While finding the sum of the numbers 1‑10, I got distracted and omitted some numbers. The sum of the remaining numbers was 50. How many different sets of numbers could I have omitted?
- The square numbers are 1, 4, 9, 16, …, and the non-square numbers are 2, 3, 5, 6, 7, 8, 10, 11, and so on. What is the 50th non-square number?
- Choose three numbers so that one number is selected in each row and each column. What’s the sum of the three numbers?

- A two-player game is played on this number rack with five rows of 10 beads. One player chooses to be Odd, the other Even. The players take turns. On each turn, a player may slide one, two, or three beads from the middle to the side of the rack. Beads moved to the side cannot be moved again. When all beads have been moved, the Odd player earns one point for each row with an odd number of beads on each side, and the Even player earns one point for each row with an even number of beads on each side. The player with the most points wins. What is the optimal strategy, and who should win?
- How many people must be present to have a probability of 50% that two of them will share a birthday?
- Insert only addition and subtraction symbols to make the following equation true:

9 8 7 6 5 4 3 2 1 = 50

- What’s the area of the square? (Inspiration from Catriona Agg, both for the puzzle and for the reduction in words.)

As promised, here are the solutions.

- Many people think you just need to adopt 4 hounds, since each hound represents 2% of the total. But that won’t work, because with each hound adopted, the total also decreases. Adopting 4 hounds would leave 45/46 ≈ 97.8%. To leave 90%, remove 40 hounds, which means that 9/10 = 90% of the remaining puppies will be hounds.
- Since 50 = 2 &*times; 25, the letters B and Y can be used. One possible word is BY. Adding an A won’t change the product, since A = 1, so another common word is BAY. Since 25 = 5 × 5, another possible word is BEE. (An obscure alternative is ABY, an archaic word meaning
*to endure*.) - 3 + 47 = 50, and 3 × 47 = 141.
- The sum 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55, which is 5 more than the result. Therefore, either 1 and 4, 2 and 3, or 5 by itself were omitted.
- Through 49, there are 7 square numbers and 42 non‑square numbers. So 49 + 8 = 57 is the 50th non‑square number.
- Doesn’t matter which three you choose, the sum will always be 50. Can you figure out how it works? And can you create a similar puzzle?
- Sorry, you’re gonna have to figure out the strategy on your own…
- This is the famous Birthday Problem, and 23 people are needed so the probability is at least 50%.
- There are many correct equations. One is 9 + 8 + 7 – 6 + 5 – 4 + 32 – 1 = 50. (The trick is to notice that at least one pair of consecutive numbers should have no sign inserted between them, and those two numbers concatenate to form a larger number.)
- The answer is 40,000 square units. You’ll need to prove to yourself that that’s correct.

Happy birthday to ME!

## Happy as L!

On Wednesday, I’ll complete my 50th trip around the Sun. To celebrate, my friend Kris sent me a card with a wonderful Roman reminder of my age:

Thanks, Kris!

Here are two relatively easy math problems associated with my birthday:

- On Wednesday, how many days old will I be?
- What are the four positive integer factors of the answer you got to Question 1? (Hint: One of the factors is the number of
**weeks**old that I’ll be.)

I recently wrote a book called One Hundred Problems Involving the Number 100. To celebrate my 50th birthday, here are ten problems involving the number 50:

- There are 50 puppies to be adopted at a shelter, and 98% of them are hounds. How many hounds must be adopted so that 90% of the remaining puppies are hounds?
- Let A = 1, B = 2, …, Z = 26. Find two common English words for which the product of the letters is 50.
- What’s the least possible product of two prime numbers with a sum of 50?
- While finding the sum of the numbers 1‑10, I got distracted and omitted some numbers. The sum of the remaining numbers was 50. How many different sets of numbers could I have omitted?
- The square numbers are 1, 4, 9, 16, …, and the non-square numbers are 2, 3, 5, 6, 7, 8, 10, 11, and so on. What is the 50th non-square number?
- Choose three numbers so that one number is selected in each row and each column. What’s the sum of the three numbers?

- A two-player game is played on this number rack with five rows of 10 beads. One player chooses to be Odd, the other Even. The players take turns. On each turn, a player may slide one, two, or three beads from the middle to the side of the rack. Beads moved to the side cannot be moved again. When all beads have been moved, the Odd player earns one point for each row with an odd number of beads on each side, and the Even player earns one point for each row with an even number of beads on each side. The player with the most points wins. What is the optimal strategy, and who should win?
- How many people must be present to have a probability of 50% that two of them will share a birthday?
- Insert only addition and subtraction symbols to make the following equation true:

9 8 7 6 5 4 3 2 1 = 50

- What’s the area of the square? (Inspiration from Catriona Agg, both for the puzzle and for the reduction in words.)

Answers will be posted on my birthday — St. Patrick’s Day! Stop back on Wednesday!

## Change the Vowel

The following puzzle contains a clue within each clue. The answer, of course, fits the clue, but each answer is also one of the words within the clue with its vowel sound changed. For instance, the clue “the distance from the top of your hat to the sole of your shoes” contains the word *hat*, and if you change the vowel sound from a short *a* to a long *i*, you get *height*, which fits the description.

As always, there’s a catch. Every answer to the clues below is a mathy word.

- The number of permutations of three different colored socks.

- This value is the same for 3 + 1 and 2 + 2.
- Do this with two odd numbers and you’ll get an even number.
- Three hours before noon.
- You may fail to correctly expand (
*a*+*b*)(*c*+*d*) if you don’t remember this mnemonic. - One represents this fractional portion of toes on the hoof of a deer.
- The last element of a data set arranged in descending order.
- The measure of central tendency made from the most common data points.
- The number of contestants ahead of third place if there’s a tie for first place.
- If the leader is forced to drop out of a race, the runner-up takes over this place.
- The square root is needed to calculate the length of the hypotenuse in this type of triangle.
- The graph of the declination of the Sun during a year can be approximated by this type of curve.

Answers

- six

- sum
- add
- nine
- FOIL
- half
- least
- mode
- two
- first
- right
- sine

## 100 Problems for the 100th Day of School

In May 2020, I delivered a webinar titled *One-Hundred Problems Involving the Number 100*. Every problem included a problem that somehow used the number 100, maybe as the number of terms in a sequence, the length of a hypotenuse in inches, or the number of digits written on a whiteboard. At the end of the webinar, NCTM President Trena Wilkerson challenged me to create a collection of 100 problems for which the answer is always 100.

So, I did.

My process was simple. I just wrote problem after problem with little concern for topic or grade level. Some of the problems were good; others were not. Some of the problems were difficult; others were easy. Some of the problems required knowledge of esoteric math concepts; others required nothing more than the ability to add and subtract. But I wrote 100 problems, then I reviewed them and deleted those that weren’t good enough. Then I wrote some more, and cut some more, and so forth, until I finally had a collection of 100 problems that were worthy.

And I’m going to share all of them with you in just a minute. But first, a math problem for which the answer is** not** 100.

As I said, I wrote the problems as they came to me, not necessarily in the order that I’d want to present them. But to keep track of things, I numbered the problems 1‑100. Since they were in the wrong order, I had to rearrange them, meaning that Problem 92 in the draft version eventually became Problem 1 in the final collection; Problem 37 became Problem 2; Problem 1 became Problem 3; and so on. You get the idea. So, the question…

You have a collection of 100 items numbered 1‑100, but the items are out of order. When you arrange the items in the correct order, how many would you expect to be labeled correctly? (Less generically, how many of my problems had the same problem number in the draft version and the final collection?)

The solution to that problem is more beautiful than I would have initially guessed. Have fun with it.

Without further ado, here is the collection:

**Problems with 100 as the Answer**

My goal was to release these problems in time for the 100th day of school, which most schools celebrate in late January or early February. I hope this collection reaches you in time. And I present the problems one per page, so you can decide which one(s) you’d like to use with your students. If you teach algebra, then perhaps you’ll print and share Problems 46 and 53; if you teach third grade, perhaps Problem 2 will be more appropriate. But the problems cover a wide range of topics and difficulty levels, so feel free to use whichever ones you like. (Be forewarned, though. The answer to **every** problem is 100, so unless your students are absolutely terrible at identifying patterns, you probably won’t want to share every problem with them. At least, not at the same time. I’m sharing this collection in time for the 100th day of school, but feel free to use any problem at any time.)

My favorite problem in the collection? I like Problem 47:

Above the bottom row, each number in a square is the sum of the two numbers below it. What value should replace the question mark?

Feel free to let me know if you or your students have a favorite.

p.s. – Bonus points if you can identify the origin of the 100 in the image at the top of this post.

## My 0.04 Seconds of Fame

In 2017, I attended the International KenKen Championship and filmed the final round, which I posted previously on this blog. But filmmakers Louis Cancel, Chris Flaherty, and Daniel Sullivan were there that day, too, and their cameras were significantly more sophisticated than my Samsung S8. Their footage of the competition, coupled with myriad interviews of competitors, organizers, and the inventor of KenKen himself, Tetsuya Miyamoto, has resulted in a new documentary, * Miyamoto and the Machine*, recently released by

*The New Yorker*. It tells the story of KenKen’s origins and attempts to answer the question, “Can a computer make puzzles as beautiful as those created by humans?”

Many aspects of the film will appeal to kenthusiasts, but my favorite moment occurs at 17:14. Competitor Ellie Grueskin is competing in the finals, and just over Ellie’s left shoulder is a barely visible, occasionally funny, middle-aged math guy holding — wait for it — a Samsung S8!

Yep, that’s me. I’m a star!

You have to ask yourself, what kind of monster would author such a shamelessly self-promotional post and not even provide one KenKen puzzle for the reader to enjoy? Definitely not me, so here you go.

After you solve the puzzle, definitely watch * Miyamoto and the Machine*. It’s 25 minutes well spent.

## A Pattern Puzzle for the New Year

Over at the Visual Patterns site, the directions state that if you click on a pattern, you’ll get to see the number of objects in the 43rd step. Why 43? I assumed that it had to do with Fawn Nguyen being a fan of Troy Polamalu — which, as far as I’m concerned, would be just one more reason to have an infinite amount of respect for her — but when I asked about it, Fawn explained that 43 was chosen as…

…a random number that was farther down the step number to prevent students from finding the number of objects recursively, but not too far.

This explanation sits well with my beliefs. In my book *One-Hundred Problems Involving the Number 100*, I stated that it’s appropriate to ask students to find the 100th term in a sequence because 100 is “big enough to exhilarate, but not so big as to intimidate.” The same could be said about 43.

Following Fawn’s lead, here’s a problem to get you in the spirit for the new year. Feel free to share this problem with your students on or near January 1.

*How many squares would be in the 43rd element of this sequence?*

Coincidentally, I shared this sequence with Fawn, and it now appears as #392 on the Visual Patterns site.

Speaking of sequences, here’s my favorite infinite sequence joke.

Infinitely many mathematicians walk into a bar. The first says, “I’ll have a beer.” The second says, “I’ll have half a beer.” The third says, “I’ll have a quarter of a beer.” They continue like this, each one ordering half as much as the last. The barman stops them and pours two beers. One of the mathematicians says, “That’s it? That’s not enough for all of us!” The bartender replies, “C’mon, folks. Know your limits.”

For fun, figure out how much beer the 43rd mathematician asked for.

And as a little more fun, guess the value of all the coins in the glass below. As a hint, there are the same number of quarters, dimes, and nickels, but three times as many pennies as dimes. (Said another way, Q:D:N:P::1:1:1:3.)

If you think about it a little, you’ll realize the answer without doing any computation.

Happy New Year!

## Mathy Zoom Backgrounds

Do you seek the admiration of your colleagues or the respect of your students?

Do you wish to create the illusion that you’re funny and cool?

Do you long to be the envy of your virtual social circle?

Unfortunately, you’re reading a math jokes blog, which means there may not be much hope for you. But a possible start may be to download some of the math joke backgrounds below for your next online meeting. I’ve been using them for the past few weeks, and I don’t think it’d be an overstatement to say that I’m now the envy of the internet. I mean, I’ve got a face for radio, but you have to admit that I look pretty fantastic when there’s a math poem above my head and equations on either side of it:

And guess what? You can look that cool, too!

To use any of the images below, simply right click and “Save Image As…,” then install them as virtual backgrounds (Zoom, Google Meet). If you’d like a better look at any of them before deciding if they’re worth valuable memory on your laptop, just click on an image to open it full screen.

## KenKen 12 Puzzle for 12/12

Today is the twelfth day of the twelfth month, and in honor of the date, here’s a 4 × 4 KenKen puzzle that has 12 as the target number in each cage. The entire puzzle has only four cages, and it only uses addition and multiplication. Have at it!

But a post with just a KenKen puzzle isn’t much of a post, especially on a math jokes blog. So let’s consider some jokes that have to do with the association between 12 and a dozen. The following are some mathematical insults you can use if you’re playing the dozens.

Yo momma is so fat, she’s proof that the universe is expanding exponentially.

The shortest distance between two points is around yo momma’s ass.

Yo momma is so fat, her volume is an improper integral.

Yo momma is so crazy, when she received a can of Pepsi from the vending machine, she started jumping up and down, yelling, “I won! I won!”

Yo momma is so dumb, she thinks convex are inmates locked in a prism.

Yo momma is so infinitely fat, she can eat as much as she wants and not gain any weight.

Yo momma thinks cosine is what she does for a loan.

Yo momma is so dumb, she sleeps with a ruler to keep track of how long she sleeps.

Yo momma is so fat, she took geometry because she heard there was gonna be π.

Yo momma is so fat, the ratio of her circumference to diameter is 4.

Yo momma is so fat, in a love triangle she’d be the hypotenuse.

Yo momma thinks *coincide* is what you should do when it’s raining.

The integral of your mom is fat plus a constant, where the constant is equal to more fat.

Yo momma is so dumb, she doesn’t know the difference between a doughnut and a coffee cup.

Yo momma is so dumb, she thinks crossing a mosquito and a mountain climber yields |mosquito| × |mountain climber| × sin(θ).

The derivative of yo momma is strictly positive.

Yo momma is so dumb, she serves beer in Klein bottles.

Yo momma is so dumb, she thinks that if two people go into a hotel and three come out, the first two must have pro-created.

Yo momma is so dumb, she can’t even solve a second‑order non‑homogeneous differential equation.

Yo momma is so fat, her dress size requires an exponent.

The limit of yo momma’s ass tends to infinity.

Yo momma is so fat, when she steps on the scale, it displays π without a decimal point.

Yo momma’s muscle-to-fat ratio can only be explained by irrational complex numbers.

Yo momma is so ugly, Pythagoras wouldn’t touch her with a 3-4-5 triangle.

## Guess the Graph

The bar graph below was created because of a recent discussion with my wife. The title and axis labels have been removed. **Can you identify the data set used to create the graph?** I’ll give you some hints:

- The data set contains 32 elements.
- It’s based on a real-world phenomenon from this year.
- The middle five categories account for 81% of the data.
- The special points marked by A, B, and C won’t help you identify the data set, but they will be discussed below.

Got a guess?

No clue? Okay, one more hint:

- The vertical axis represents “Teams.”

Still not sure? Final hints:

- Point A represents the lowly J-E-T-S, who are currently winless.
- The region outlined by B shows that 26 teams have from 3 to 7 wins.
- Point C on the graph represents my Pittsburgh Steelers, whose record is a perfect 10‑0. (It’s my hope that I’ll still be able to gloat on Friday morning, after the Steelers host the Ravens on Thanksgiving night.)

This graph was generated while discussing the current standings in the NFL with my wife, who speculated that there seemed to be a lot of really good teams and a lot of really bad teams this year. The horizontal axis represents the number of wins. As it turns out, the distribution above is somewhat typical at this point in the season. At the end of most seasons, about 2/3 of the teams finish a 16-game season with 5 to 10 wins. It may be a little unusual that there are 8 teams with 7 wins, but it’s not statistically cray-cray.

If you’ve read this far, then you may enjoy these other math-related football trivia questions:

- Describe two ways in which an NFL game can end with a score of 2‑0.
- What’s the greatest score that cannot be attained by scoring only touchdowns (7 points) and field goals (3 points)?
- Express the ratio of width:length of a football field. For length, include the end zones.
- What are the only positions allowed to wear single-digit uniform numbers?
- During a typical broadcast of an NFL game, approximately what percent of the time is spent actually playing football (as opposed to commercials, half time, or just milling around between snaps)?

Happy Drinksgiving! And, go Stillers!

—

Answers

- A game can end 2‑0 if one team scores a safety and the other team doesn’t score at all. It can also end 2‑0 if one team forfeits before either team has scored, by league rule. (In high school and college, a forfeit is officially recorded as a 1‑0 loss.)
- 11 points. Any point total above that is (theoretically) possible. Below that, it’s not possible to score 1, 2, 4, 5, or 8 points.
- A field is 53 1/3 yards wide and 120 yards long. In feet, that’s 160:360, which can be reduced to 4:9.
- Quarterbacks and kickers.
- According to several analyses, 11 minutes of a three-hour broadcast is spent actually playing. That’s about 3%. Sheesh.