Posts filed under ‘Uncategorized’

Math Puzzles with Letters

This week on the NPR Sunday Puzzle, host Will Shortz offered the following challenge:

Name a famous city in ten letters that contains an s. Drop the s. Then assign the remaining nine letters their standard value in the alphabet — A = 1, B = 2, C = 3, etc. The total value of the nine letters is only 25. What city is it?

It’s not much of a spoiler to note that the average value of those nine letters must be less than three, since their sum “is only 25.” Consequently, a lot of those letters must occur at the beginning of the alphabet and — if eight of them were a‘s — there would be no letters later than q in the name of the city. But that’s as much as I’ll say; you can solve the puzzle on your own. (When you do, you can submit your answer for a chance to play next week’s on-air puzzle live with Will Shortz.)

Mathematician Harold Reiter uses a similar problem with elementary school students. Using the same idea — that each letter has a value (in cents) equal to its position in the alphabet — he asks students to find a dollar word, that is, a word whose letters have a sum of 100. As it turns out, there are many. Based on a nonexhaustive search, there are at least 3,500 dollar words, and likely a whole lot more. In a quick perusal of the list, one word jumped out: oxygon. Nope, that’s not a typo. It’s an archaic term meaning “a triangle with three acute angles.”

All of this talk of letters reminds me of my favorite puzzle, which I call Product Values. Using the same scheme — that is, A = 1, B = 2, C = 3, etc. — find the product value of a word by multiplying the values of the letters. So, for instance, cat has a product value of 3 × 1 × 20 = 60. How many words can you find that have a product value of 100? Based on the ENABLE word list, there are nine. (If you need some help, you can use the Product Value Calculator at www.mathjokes4mathyfolks.com.)

To end this post, a few math jokes that involve letters:

And Satan sayeth, “Let’s put the alphabet in math.” Bwa-ha-ha-ha-ha.

Romans had no trouble with algebra, because X was always equal to 10.

June 1, 2021 at 6:36 am 4 comments

Brood X Anniversary

The Brood X cicadas have erupted again after 17 years, which can mean only one thing: it must be our 17th wedding anniversary.

You see, my wife and I were married in 2004, the last time the Brood X cicadas surfaced. Despite our poor choice of date — honestly, who plans an outdoor wedding during peak cicada time? — we were unaffected. Hundreds of cicadas were chirping away about a mile from the venue, but only two made an appearance at the event: one was seen crawling across the stage in front of the band, and the other was sunning itself on the concrete wall of a water fountain in the garden.

My best man and his wife recently sent us the following photo:

The accompanying text read, “They’re ba-a-a-a-ack! It must be your 17th anniversary!”

At dinner on Sunday night, my wife handed me a card with this image on the front:

She knows me well, and she knew a Venn diagram (no relation) would resonate. Of course, I couldn’t help but wonder why the area of overlap was so small! After 17 years, shouldn’t the image look more like this?

Was she trying to tell me something? Luckily, the message inside gave a clear indication that I had nothing to worry about. (No, I won’t elaborate as to what it said. Use your imagination. Nice boys don’t kiss and tell.)

Venn diagrams are useful for expressing relationships between two or more things. This one appeared on Twitter at least five years ago, but I only discovered it recently.

In my humble opinion, the king of Venn diagrams is Demetri Martin, from whom we get this wonderful comparison of ants, bears, and people:

My personal contribution to the genre stems from a realization about the disciplines that claim April as their national month: math, poetry, and humor:

Finally, a Venn diagram tautology.

May 20, 2021 at 1:57 am Leave a comment

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:

Plaques given to me by MathCounts,
because I have very talented sons.

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  
XWXZ
XWYZ
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.

March 25, 2021 at 12:37 pm Leave a comment

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:

  1. 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?
  2. Let A = 1, B = 2, …, Z = 26. Find two common English words for which the product of the letters is 50.
  3. What’s the least possible product of two prime numbers with a sum of 50?
  4. 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?
  5. 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?
  6. Choose three numbers so that one number is selected in each row and each column. What’s the sum of the three numbers?

  1. 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?
  2. How many people must be present to have a probability of 50% that two of them will share a birthday?
  3. Insert only addition and subtraction symbols to make the following equation true:

9    8    7    6    5    4    3    2    1 = 50

  1. 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.

  1. 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.
  2. 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. 3 + 47 = 50, and 3 × 47 = 141.
  4. 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.
  5. Through 49, there are 7 square numbers and 42 non‑square numbers. So 49 + 8 = 57 is the 50th non‑square number.
  6. 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?
  7. Sorry, you’re gonna have to figure out the strategy on your own…
  8. This is the famous Birthday Problem, and 23 people are needed so the probability is at least 50%.
  9. 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.)
  10. The answer is 40,000 square units. You’ll need to prove to yourself that that’s correct.

Happy birthday to ME!

March 17, 2021 at 8:42 am 3 comments

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:

  1. On Wednesday, how many days old will I be?
  2. 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:

  1. 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?
  2. Let A = 1, B = 2, …, Z = 26. Find two common English words for which the product of the letters is 50.
  3. What’s the least possible product of two prime numbers with a sum of 50?
  4. 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?
  5. 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?
  6. Choose three numbers so that one number is selected in each row and each column. What’s the sum of the three numbers?

  1. 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?
  2. How many people must be present to have a probability of 50% that two of them will share a birthday?
  3. Insert only addition and subtraction symbols to make the following equation true:

9    8    7    6    5    4    3    2    1 = 50

  1. 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!

March 15, 2021 at 5:11 am Leave a comment

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.

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

Answers

  1. six
  2. sum
  3. add
  4. nine
  5. FOIL
  6. half
  7. least
  8. mode
  9. two
  10. first
  11. right
  12. sine

January 23, 2021 at 6:24 am 1 comment

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.

January 13, 2021 at 3:30 am Leave a comment

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.

January 9, 2021 at 8:15 am Leave a comment

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!

December 28, 2020 at 8:01 am Leave a comment

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.

Opinion Minus Pi

Trig Tank

Binary

Root Beer

Punch Line

Pi and E

Pentagon, Hexagon, Oregon

Tom Swiftie

Graph Paper

Complex Person

6 Afraid of 7

Math and Coffee

 

 

December 21, 2020 at 6:21 am Leave a comment

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About MJ4MF

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.

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