## Archive for March, 2021

### 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:

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 XWXZXWYZ 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**.

### 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:

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

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

9 8 7 6 5 4 3 2 1 = 50

- 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.

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

Happy birthday to ME!

### 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:

- On Wednesday, how many days old will I be?
- 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:

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

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

9 8 7 6 5 4 3 2 1 = 50

- 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!