Posts tagged ‘competition’

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

Wu, California Earn MathCounts Titles

Today, I had the privilege of attending the MathCounts National Competition. I served as a volunteer in the scoring room, and I had the pleasure of scoring the competitions of the winning team. Congratulations to the California team of Celine Liang, Sean Shi, Andrew He and Alex Hong, who scored an amazing 59.5 (out of 66) on the competition.

MathCounts Logo

You may be thinking that 90.1% correct is not very impressive. With the current level of grade inflation, I suppose that’s a reasonable reaction. But given that these students solve 48 non‑trivial questions (see examples below) in under two hours, it’s an amazing feat. Given twice as much time, most of us would be lucky to answer half of them correctly.

During the Target Round of the competition, students are given six minutes to solve a pair of problems. My favorite problem on this year’s competition was the last question in the Target Round:

How many positive integers less than 2011 cannot be expressed as the difference of the squares of two positive integers?

It’s a good question. You might like to try solving it. But don’t be too discouraged if it takes you more than six minutes… after all, no one expects you to be as smart as a middle school student.

One of the California competitors, Sean Shi, scored well enough in the written competition to qualify for the Countdown Round. The Countdown Round is an NCAA‑style bracket for the top twelve competitors. All students who qualify for the Countdown Round are introduced to the audience, and the following story was told about Sean:

Sean’s younger brother is two years younger than Sean. When Sean was 7 and his brother was 5, his brother told their mom, “You’re the best mommy in the whole world!” Already a mathematical prodigy, Sean replied, “Actually, the chance of that being true is very low.” Sean explained, “No offense, mom! I’m just being mathematical!”

Another competitor in today’s Countdown Round, Shyam Narayanan of Kansas, finished among the top four in last year’s national competition. As a result, he was invited to meet President Obama in an Oval Office ceremony. During the visit, Shyam asked the President a question that he said he had never been asked before.

Since the Oval Office is an ellipse, where are its foci?

Although Obama didn’t know the answer, the mathletes in attendance were able to help him figure it out.

Though Shi and Narayanan made respectable showings in the Countdown Round, it was Scott Wu from Louisiana who prevailed and earned the title of national champion. Wu, who’s older brother Neil Wu was the 2005 MathCounts National Champion, defeated Yang Liu 4‑1 in the finals. Wu locked up the title by correctly solving a rather odd question about digits.

It takes 180 digits to write all of the two-digit positive integers. How many of the digits are odd?

In the Countdown Round, students only have 45 seconds to answer each question — though typically, they answer the questions much faster. So how do you come up with an answer to this question in just a few seconds?

First, realize that exactly half of the two‑digit numbers have an odd units digit. Then, notice that 50 numbers have an odd tens digit, but only 40 have an even tens digit. Consequently, of the 180 digits required, there will be 10 more odd digits than even. Hence, 95 digits are odd, and 85 digits are even.

May 6, 2011 at 6:25 pm 1 comment


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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