Posts tagged ‘problems’

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!

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

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

One-Hundred Problems Involving the Number 100

Although the following joke appears in Math Jokes 4 Mathy Folks —

Why was the math book sad?
Because it had so many problems.

— I’ve often contended that it isn’t true. Math books aren’t sad because they have too many problems. They’re sad because they have too many exercises.

But my forthcoming book isn’t the least bit melancholy, because it contains a multitude of honest-to-goodness, classroom-tested, student-approved, 100% legit math problems — a century of them, in fact, as implied by the title.

One-Hundred Problems Involving the Number 100 by G. Patrick Vennebush. Available now for pre-order from NCTM.

Disclaimer: The title is a lie. The book actually contains 101 problems. I was so excited, I just couldn’t stop myself when I got to 100. But don’t you worry; there’s no charge for that extra 1%.

As a sample, here are four problems from the book. To experience a fifth problem, register for an NCTM Author Panel Talk on Wednesday, October 7, 7:00 p.m. ET, when Marian Small, Roger Day, and I will be discussing rich tasks and sharing samples from each of our new books. The webinar will be moderated by NCTM Board Member Beth Kobett. Hope to see you there!

October 2, 2020 at 9:08 am Leave a comment

Grid with 100 Paths

Due to current circumstances, the National Council of Teachers of Mathematics (NCTM) had to cancel their Centennial Annual Meeting and Exposition, which was to be held April 1-4 in Chicago. As a replacement, though, they presented an amazing gift to the math education community — 100 free webinars led by selected speakers from the Chicago program. Dubbed 100 Days of Professional Learning, these webinars are to be held on select days from April through October.

As part of the 100 Days, I presented “100 Problems Involving the Number 100” on May 14. To celebrate NCTM’s 100th Anniversary, I collected or created 100 problems, each of which included the number 100. One of my favorites, Dog Days, looked like this:

From this information alone, what could you determine?

At the end of the webinar, the conversation continued “backstage” with several members of NCTM staff, NCTM President Trena Wilkerson, and me. During that conversation, Trena made the outlandish suggestion,

Now we need a collection of 100 problems for which the answer is always 100.

I had just finished preparing a webinar with 100 problems, and now she was asking for another 100 problems. But never one to shy away from a challenge, I began to think about what kinds of problems might be included in such a collection. One type that came to mind was path-counting problems, like this one:

Moving only north or east on the segments in the diagram, how many distinct paths are possible from A to B?

That particular grid, measuring just 3 × 4, has fewer than 100 distinct paths from A to B. (How many paths, exactly? That’s left as an exercise for the reader.) What got me excited, though, was wondering if there were any grids that have exactly 100 paths — and hence providing 1% of the content for the collection that Trena requested.

As it turns out, there are no unmodified m × n grids that have 100 distinct paths. But what if some segments were removed? For instance, what if one of the middle vertical segments were discarded from a 4 × 6 grid, as shown below? How many distinct paths from A to B would there be?

As it turns out, a lot more than 100. (Again, finding the exact number is an exercise I’ll leave for you. If you need help, Richard Rusczyk from Art of Problem Solving has a video showing how to count paths on a grid.)

So, this is where I leave you:

Can you create a grid with some segments removed that will have exactly 100 distinct paths?

Have fun! Good luck!

As for the webinar, which lasted only 60 minutes, there wasn’t nearly enough time to cover 100 problems, but we had fun with 5 of them.

If you missed the webinar, you can hear the discussion about the Dog Days problem and 4 others, as well as get a PDF of all 100 problems, via the links below.

Enjoy!

May 28, 2020 at 9:36 am 3 comments

Math of the Rundetaarn

RundetaarnAs we were exiting the Rundetaarn (“Round Tower”) in Copenhagen, Denmark, I noticed a man wearing a shirt with the following quotation:

Find what you love, and let it kill you.

The only problem is that the shirt attributed the quotation to poet Charles Bukowski, when apparently it should have been attributed to humorist Kinky Friedman. For what it’s worth, my favorite Friedman quote is, “I just want Texas to be number one in something other than executions, toll roads, and property taxes.” But this ain’t a post about Kinky Friedman, or even Charles Bukowski. So, allow me to pull off the sidewalk and get back on the boulevard.

Whoever said it, the quotation hit me as drastically appropriate. I suspect that math will someday kill me… likely as I cross the street while playing KenKen on my phone, oblivious to an oncoming truck. As I exited the Rundetaarn, I was thinking about all the math that I had seen inside — much of which, I suspect, would not have been seen by many of the other tourists.

The Rundetaarn, completed in 1642, is known for the 7.5-turn helical ramp that visitors can walk to the top of the tower and, coincidentally, to one helluva view of the city. Rundetaarn Cross SectionThat leads to Question #1.

Along the outer wall of the tower, the winding corridor has a length of 210 meters, climbing 3.74 meters per turn. What is the (inside) diameter of the tower?

Above Trinitatis Church is a gift shop that is accessible from the Rundetaarn’s spiral corridor. The following clock was hanging on the wall in that little shop:

Rundetaarn clock

I have no idea who the bust is, but the clock leads to Question #2.

What sequence of geometric transformations were required to convert a regular clock into this clock?

And to Question #3.

Do the hands on this clock spin clockwise or counterclockwise?

And to Question #3a.

What is the “error” on the clock?

A privy accessible from the spiral corridor in the Rundetaarn has been preserved like a museum exhibit. Sadly, I have no picture of it to share, but a sign next to the privy implied that the feces deposited by a friar would fall 12 meters into the pit below.

That leads to Question #4.

What is the terminal velocity of a deposit when it reaches the bottom of the pit? (Or should that be “turd-minal velocity”?)

The first respondent to correctly answer all of these questions will earn inalienable bragging rights for perpetuity.

August 11, 2016 at 8:31 am Leave a comment

Common Problems

This is pretty cool. What is the value of the following expression?

4 + π5)1/6

This reminds me of a series of problems that I call Maximum Mileage:

  1. What is the maximum possible product of two positive integers whose sum is 100?
  2. What is the maximum possible product of two prime numbers whose sum is 100?
  3. A set of positive integers has a sum of 100. What is the maximum possible product of the numbers in this set?
  4. A set of positive numbers has a sum of 100. What is the maximum possible product of the numbers in this set?

Bonus Question: Why does the expression at the beginning of this post remind me of that series of problems?

The following sentence might be a hint:

It is fortuitous that both conundrums incur a commonality of solution.

November 30, 2013 at 8:04 am Leave a comment

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!)

MathCounts

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:

MathCounts QWC Shirt

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:

Marble Bags

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?

Divided 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:

Square Divided Into 8 Squares

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%.

May 3, 2011 at 7:48 am 2 comments

Welcome to 2011

An original problem to kick off the new year:

Place three operators (+, –, ×, ÷) between the digits below to make an expression equal to 2011.

1  2  3  4  5  6  7

For instance, using three + signs, you could make the expression 123 + 45 + 6 + 7, which is equal to 181.

A more standard problem to begin any new year is this:

Combine the digits of the year (in this case, 2, 0, 1, 1) to create each of the numbers 1–100. Extra credit if you can keep the digits in the same order as they appear in the year. For instance, 20 – 1 + 1 = 1 would be a way to make 1 with the digits in order, and 10 – 2 + 1 = 9 would be a way to make 9 with the digits out of order. 

I hope that 2011 brings you peace, love, happiness, and lots of problems!

January 1, 2011 at 12:00 am 2 comments

Movie: Fermat’s Room

I’m not sure that the indie film Fermat’s Room deserved to win four awards or deserved a nomination for “Best Film” at the Sitges International Film Festival, but it’s got enough gems to keep mathy folks entertained for almost 90 minutes. 

Take, for instance, this great line:

The more you study logic, the more you value coincidence.

In a moment, I’ll tell you about all the great math problems within the film. But first, let me tell you a little about the movie itself.

The general idea (without being a spoiler) is this: Four mathematicians are trapped in a shrinking room. Every so often, a math puzzle appears on a PDA, and they have one minute to enter the correct answer. If they take longer than a minute, the room starts shrinking — literally. Behind each wall is a hydraulic press that pushes toward the center until the correct answer is entered. While working out these riddles, there are two greater puzzles that they are attempting to solve — who would have done this, and how can they escape?

My favorite scene is when the young, brash, theoretical mathematician and the middle-aged, stoic, applied mathematician think they may have found a way to stop the hydraulic presses. “Will it work?” asks the theoretical mathematician.

“The only way to find out is to do it,” says the applied mathematician.

Upon hearing this, the young mathematician starts writing equations on a piece of paper, attempting to prove (theoretically) that their solution will work. The applied mathematician, who has already started to implement the solution, shakes the theoretical mathematician’s shoulder, as if to say, “No, really, we need to try it and see if it works, not just prove that a solution exists.”

It’s a fantastic and not-so-subtle commentary on the tension between theoretical and applied mathematicians. I laughed out loud.

But it’s got more than just great lines. It contains a treasure trove of famous math puzzles. I’ve listed several of them below — without context, so as not to spoil the movie; and without solutions, so as not to spoil your fun in solving them. Enjoy!

  1. En que orden estan los siguentes numeros? 5, 4, 2, 9, 8, 6, 7, 3, 1
    (Note: It’s a huge hint that this problem is presented in Spanish. If presented in English, the order of the numbers would be different, and the problem would read as follows: What is the order of the following numbers? 8, 5, 4, 9, 1, 7, 6, 3, 2)
  2. Three boxes contain marbles. One box contains red marbles, another contains blue marbles, and the third contains a mixture of red and blue marbles. The boxes are labeled “Red,” “Blue,” and “Mixture,” but none of the boxes contains the correct label. What is the least number of marbles you could remove to know the contents of each box?
  3. You have two egg timers, one that measures four minutes and one that measures seven minutes. How can you use them to measure exactly nine minutes?
  4. (This one’s my favorite from the movie. I originally read it in a Martin Gardner book.) A professor tells his students, “I have three daughters, and the product of their ages is 36. How old are my daughters?”
    His students work on the problem for a few minutes, then a woman in the class says, “I’m sorry, professor, but that’s not enough information to solve the problem.”
    “Ah, yes,” he says. “I should have told you that the sum of their ages is equal to my house number.”
    “I’m sorry, sir,” she says. “That is still not enough information to solve the problem.”
    The professor asks, “Will it help if I tell you that the oldest one plays piano?”
    “It will,” says the woman. “I now know the ages of your daughters.”
    Based on the information, can you determine the ages of the professor’s daughters?

July 26, 2010 at 6:54 am 7 comments


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.

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