Posts tagged ‘solution’

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

Stolen Truck Solution

Last week, I posted a modified version of Marilyn Burn’s horse problem, and I asked you to submit your answers. I received 331 responses; thanks for participating! An analysis of the submitted responses appears below, but first, a few comments and several different solutions.

My friend Jeane Joyner of Meredith College uses this problem in teacher and parent workshops. She said:

I have folks move to corners of the room — makes money, loses money, and breaks even. Then each groups selects an ambassador to go to the other groups to see if they can persuade folks to move. Fun!

Without further adieu, here is the answer: The man made $200.

Solution #1: He spent 600 + 800 = $1,400, and he received 700 + 900 = $1,600. That’s a profit of $200.

Solution #2: Assume the man started with $1,000 in his bank account. He bought the truck for $600, so he had $400 left. He then sold it for $700, so his account increased to $1,100. He bought it back for $800, so he had $300 left. When he sold it for $900, his account increased to $1,200. Since he started with $1,000 and ended with $1,200, he made a profit of $200.

Solution #3: Use a number line to show how his amount of money changed.

Truck - Number Line

After the four transactions, he is at +200 on the number line, so he made a profit of $200.

Solution #4: Some people find it confusing that he buys and sells the truck twice. It might be easier to think of him doing these transactions with two different vehicles. For instance, what if he bought a truck for $600 then sold it for $700, and then bought a car for $800 and sold it for $900? It might be easier to wrap your head around that.

Approached that way, his actions represent two separate events. The first time he bought and sold the truck, he paid $600 and sold it for $700. That’s a profit of $100. The second time, he bought it for $800 and sold it for $900. That’s another $100 profit. In total, he made $200.

After I posted the problem, a friend on Facebook asked, “Huh? How is that supposed to be hard?” Edward Early of St. Edwards College responded:

Sadly, I’ll only be surprised if there is a strong consensus for the correct answer. I’ve been teaching math too long to expect that to happen.

Of the 331 respondents, only 325 submitted usable responses. (Chalk this up to bad phrasing on my part. I asked folks to “enter a negative number if he lost money, 0 if he broke even, or a positive number if he made money.” I was meaning for people to enter the amount he lost or made, but some respondents entered a positive number that wasn’t possible in the context of the problem, such as 1 or 12. I think they thought they should enter any positive number to indicate that he made a profit. And one wiseacre responded, “a positive number.” Sorry, not even partial credit for that response!)

Of the 325 usable responses, 228 (70.1%) were correct. Answers from the other 28.9% ranged from ‑600 to 900, with 0 (28 responses) and 100 (47 responses) chosen most often. The chart below shows the distribution of incorrect responses. (Data for 200 has been removed since it overwhelms the others; its bar would be more than four times the height of the next highest bar.)

Truck Responses

The vast majority of respondents (276) were 16 years of age or older. The 49 responses from people age 15 or younger looked like this:

Age Responses Correct Responses
6 1 0
7 1 1
8 2 1
9 4 1
10 4 1
11 1 1
12 4 3
13 15 11
14 8 7
15 9 7

Interestingly, the under‑16 crowd, with 67.3% correct responses, did almost as well as the over‑16 crowd, which had 70.6% correct responses. And for the 11- to 15‑year old subset, an astounding 78.4% of the responses were correct.

From this, we can conclude that these youngsters are smarter than the rest of the population… but of course we already knew that, because teens and pre-teens know everything, right?

September 20, 2012 at 1:00 am 1 comment

Word Squares Revisited

On December 22, I posed the following puzzle in the Square Deal post:

Create a 4 × 4 grid composed of common English words (you can use the list of official Scrabble four‑letter words as a reference) such that the sum of the point values of the 16 letters is as high as possible.

In that post, I mentioned that I had created a grid worth 62 points, and many readers have asked about it. Here it is:

Z I Z Z
I D E A
Z E S T
Z A T I

 

All of the words are acceptable when playing Scrabble. While IDEA and ZEST are very common, the other two are not:

  • ZIZZ = a nap
  • ZATI = a species of macaque

I am certain that a better solution exists, but I have neither the time nor the energy to search for it. Anyone out there want to write a computer program to solve it?

** UPDATE (1/14/11): Check out the comments section from the previous post. Veky Edgar wrote a computer program and found the maximum possible score (92 points), and Scandinavian manually created a grid worth 85 points.

January 13, 2011 at 8:34 am Leave a comment

The Perfect Pack – My Solution

Here’s a question I posted a few days ago:

A standard pack of M&M’s contains pieces of six different colors. What is the probability that there will be an even number of M&M’s of each of the six colors?

As far as I’m concerned, the type of pack you select is irrelevant, as is the proportion of the colors within the pack. Even though MARS® claims a certain proportion for the mixtures at the factory, the proportion of colors varies from bag to bag. Therefore, my solution is independent of the number of M&M’s.

For each color, the number of M&M’s will be either even or odd. Consequently, the probability of having an even number of one specific color is 1/2. Since there are six different colors, then the combined probability of having an even number of every color is (1/2)6 = 1/64.

I’ve eaten way more than 64 packs of M&M’s in my life, so I’m surprised that I’ve never encountered a “perfect pack.”

September 4, 2010 at 11:57 pm 8 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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