Posts tagged ‘problems’

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.

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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.

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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