Sidhu Moose Wala Would Love This Post

[Sidhu Moose Wala – 47]

The following is a modified version of a problem that appeared in Concept Quests, a collection of problems originally written by Kim Markworth and soon to be offered by The Math Learning Center as a supplement to Bridges in Mathematic Second Edition.

I was today years old when I realized that I’ve been alive for 47 years, 47 months, 47 weeks, and 47 days. How old will I be on my next birthday?

Incidentally, the number 47 was also featured in this week’s challenge on the NPR Sunday Puzzle:

Take this equation: 14 + 116 + 68 = 47. Clearly this doesn‘t work mathematically. But it does work in a nonmathematical way. Please explain.

The number 47 is very interesting, especially if you’re a member of The 47 Society. But even if you aren’t, the number 47 has some noteworthy credentials:

• The number 47 is a safe prime, meaning it has the form 2q + 1, where q is also prime.
• The greatest number of cubes (not necessarily the same size) that cannot be rearranged to form a larger cube is 47. That is, it’s not possible to make a cube from 47 smaller, not-necessarily-the-same-size cubes, but it is possible to make a cube from 48, 49, 50, 51, or any greater number of cubes.
• 47 = 2⁵ + 5² – 2 × 5
• Lord Asano Takumi’s followers swore to avenge his death — a suicide he was obliged to commit for drawing his sword in the palace — and survives in a story known as the Legend of the 47 Ronin.
• Mexican revolutionary Pancho Villa was killed by 47 bullets.
• It takes 47 divisions of one cell to produce the number of cells in the human body; that is, there are approximately 2⁴⁷ cells in a human.
• Proposition 47 of Euclid’s Elements is the Pythagorean theorem.
• The Bible credits Jesus with 47 miracles.
• The Declaration of Independence has 47 sentences.
• There are 47 strings on a concert harp.
• The tropics of Cancer and Capricorn are located 47 degrees apart.

This fascination with the number 47 seems to have started a long time ago.

In the summer of 1964, two students at Pomona College — Laurens “Laurie” Mets and Bruce Elgin — began an extracurricular experiment to determine how often the number 47 occurred in nature. While this may seem like an odd thing to do, they weren’t the only ones; a number of Pomona students were involved in a summer program sponsored by the NIH, and many of them conducted similar experiments about the occurrence of random numbers, questioning the existence of patterns in nature. However, it was Mets’s and Elgin’s search for 47 that took on a life of its own and developed a cult following.

The experiment was to determine if 47 showed up in nature more often than other numbers. As a first step, they hypothesized that 47 would appear in two percent of all California license plates, which would be higher than would occur by random chance. According to Mets, a funny thing happened: when they looked at a bunch of license plates and counted the occurrence of 47, they found their hypothesis to be true. According to Elgin, however, the license plate experiment was a failure; but, within the next week, he saw a Rolaids commercial claiming that Rolaids absorbs up to 47% of its weight in excess acid. That sparked a desire to see where else 47 might occur… and before long, people all over campus were counting everything to look for 47s.

As part of that same summer program, statistician Donald Bentley gave a talk in which he stated that any number could be shown to be equal to any other number. In his (invalid) proof, he chose to show that every number was equal to — you guessed it — 47. How did he prove it?

Consider an isosceles right triangle for which the base is 47 units, and the congruent legs are each 37 units. Now, connect the midpoints of the congruent sides to the midpoint of the base, as shown in Figure 1. The combined lengths of the red segments in Figure 1 is the same as the sum of the lengths of the congruent sides, 74 units. Then repeat, connecting the midpoints of these smaller triangles, as shown in Figure 2. Again, the combined lengths of the red segments is 74 units. Then do this again and again and again — all the way to infinity — and eventually the triangles will get so small that the red segments will appear to be a straight-line segment that overlaps the base of the triangle, as shown in Figure 3. The length of the red segment is 74 units, and length of the base is 47 units, so 74 = 47. By similar (incorrect) reasoning, it would be possible to show that any number is equal to 47.

Although Dr. Bentley’s proof was meant as a classroom lesson on statistical computing, leave it to the 47-hunters to accept the proof and believe that all numbers equal 47.

(If you’d like to know more about this sordid history of 47, you can read The Mystery of 47 at the Pomona College website.)

Anyway, where was I? Oh, yes… I’ve been alive for 47 years, 47 months, 47 weeks, and 47 days.

Mathematically, what I find most interesting is the general case. That is, if I’ve been alive for n years, n months, n weeks, and n days, how old will I be on my next birthday?

For the problem I presented at the top of this post, n = 47, and the answer is 52 years; that is, I’ll turn 52 on my next birthday. But if n = 48, I won’t turn 53 on my next birthday; I’ll turn 54. That’s right, it skips a year. This happens periodically, and with a little help from the Date Calculator at www.timeanddate.com, I was able to generate this beautiful — albeit incomplete — graph:

 If you’re interested in seeing the data, check it out at https://www.desmos.com/calculator/u43gz6ahyi.

Something really wacky happens when n = 38. I was born on March 17, 1971, and the date 38 years, 38 months, 38 weeks, and 38 days from my date of birth was exactly 42 years later: March 17, 2013. Whoa.

Who knows what any of this means? Perhaps this implies that 38, not 47, is a magical number. In any case, I hope you have fun playing with this scenario and seeing what you can discover.

If that’s not your thing, though, here are some time-and-date jokes for you:

How many seconds are in a year?
Twelve. January 2nd, February 2nd, March 2nd, …

Did you hear about the hungry clock?
It went back four seconds.

Traditional calendars are for people who are week-minded.

I was fired from my job at the calendar factory. My boss was mad that I took a few days off.

Math Problems for 2023

Happy New Year!

I’m feeling lucky about 2023… but perhaps that’s because 7⁷ mod 7! = 2023.

But I’m also feeling lucky because there are many interesting problems that involve the number 2023. I’m a little late in getting this post out, but all of the following problems attempt to use the number 2023 in some interesting way. Thanks to Professor Harold Reiter from UNC‑Charlotte, who supplied the ideas for the first two problems; the rest are MJ4MF originals. Enjoy!

• For 2023, the sum of the digits of the year, times the square of the sum of the squares of the digits, is equal to the year itself. Yeah, that’s a mouthful; more concisely, (2 + 0 + 2 + 3) × (2² + 0² + 2² + 3²)² = 2023. Can you find the only other four-digit number abcd with the property that (a + b + c + d) × (a² + b² + c² + d²)² = abcd?
• If you place one dot inside an equilateral triangle and connect it to each vertex, you get three non-overlapping triangles. Similarly, if you place two dots in an equilateral triangle — and no subset contains three collinear dots — you can connect the dots to form five non-overlapping triangles. How many non-collinear dots must you place inside the triangle to get 2023 non-overlapping triangles?
• The sum of the digits of 2023 is 2 + 0 + 2 + 3 = 7, and 7 is a factor of 2023. For how many numbers in the 2020s is the sum of the digits a factor of the number?
• In how many ways can a 4 x 4 grid be covered with monominos and L-shaped triominos? One such covering is shown below.

• What is the sum of 1 + 3 + 5 + 7 + ··· + 2023?
• How long would it take you to count to 2023?
• Using only common mathematical symbols and operations and the digits 2, 0, 2, and 3, make an expression that is exactly equal to 100. (Bonus: make an expression using the four digits in order.)
• All possible four-digit numbers that can be made with the digits 2, 0, 2, and 3 are formed and arranged in ascending order. What is the first number in the list?
• Place addition or subtraction symbols between the cubes below to create a true equation:

9³     8³     7³     6³     5³     4³     3³     2³     1³ = 2023

• Find a fraction with the following decimal equivalent.

• How many positive integer factors does 2023 have?
• Create a 4 × 4 magic square in which the sum of each row, column, and diagonal is 2023.
• What is the units digit of 2023²⁰²³?
• What is the value of the following expression, if x + 1/x = 2?

• Each dimension of a rectangular box is an integer number of inches. The volume of the box is 2023 in³. What is the minimum possible surface area of the box?
• What is the maximum possible product for a set of positive integers that have a sum of 2023?

A Pattern Puzzle for the New Year

Over at the Visual Patterns site, the directions state that if you click on a pattern, you’ll get to see the number of objects in the 43rd step. Why 43? I assumed that it had to do with Fawn Nguyen being a fan of Troy Polamalu — which, as far as I’m concerned, would be just one more reason to have an infinite amount of respect for her — but when I asked about it, Fawn explained that 43 was chosen as…

…a random number that was farther down the step number to prevent students from finding the number of objects recursively, but not too far. 

This explanation sits well with my beliefs. In my book One-Hundred Problems Involving the Number 100, I stated that it’s appropriate to ask students to find the 100th term in a sequence because 100 is “big enough to exhilarate, but not so big as to intimidate.” The same could be said about 43.

Following Fawn’s lead, here’s a problem to get you in the spirit for the new year. Feel free to share this problem with your students on or near January 1.

How many squares would be in the 43rd element of this sequence?

Coincidentally, I shared this sequence with Fawn, and it now appears as #392 on the Visual Patterns site.

Speaking of sequences, here’s my favorite infinite sequence joke.

Infinitely many mathematicians walk into a bar. The first says, “I’ll have a beer.” The second says, “I’ll have half a beer.” The third says, “I’ll have a quarter of a beer.” They continue like this, each one ordering half as much as the last. The barman stops them and pours two beers. One of the mathematicians says, “That’s it? That’s not enough for all of us!” The bartender replies, “C’mon, folks. Know your limits.”

For fun, figure out how much beer the 43rd mathematician asked for.

And as a little more fun, guess the value of all the coins in the glass below. As a hint, there are the same number of quarters, dimes, and nickels, but three times as many pennies as dimes. (Said another way, Q:D:N:P::1:1:1:3.)

If you think about it a little, you’ll realize the answer without doing any computation.

Happy New Year!

Covering 100 Squares

There’s an old math joke that says math books are sad because they have too many problems. But I disagree; I believe that most math books — and, in particular, textbooks — are sad because they have too many exercises.

To try to release more true problems into the wild, I recently wrote a book called One-Hundred Problems Involving the Number 100. Published by NCTM, the book contains problems, suggestions for classroom use, and solutions. Some of the problems are old chestnuts, such as, “What is 1 + 2 + 3 + ⋯ + 100?” (Problem 21: Gauss and Check). Others are complete originals, and occasionally a little silly, such as, “If all the positive integers from 1 to 100 were spelled out, how many letters would be used?” (Problem 1: Spell It Out). But my favorite problem in the book, which I’ve shared twice during NCTM’s 100 Days of Professional Development (May 14, Oct 7; login required), is Problem 100: Covering with Squares.

As shown below, a square grid with 100 smaller squares can be covered by 100 squares (each measuring 1 × 1), by 25 squares (each measuring 2 × 2), or by 13 squares (one 6 × 6, two 4 × 4, two 3 × 3, two 2 × 2, and six 1 × 1).

Find all values of 0 < n < 100 for which it is impossible to cover a 10 × 10 grid with n squares of integer side length.

I’ll admit, I’m not ecstatic about the phrasing of the question at the end. The symbolic representation makes it succinct, sure, but this problem can be investigated by students who may be too young to understand that notation. It might be better to just ask, “Can you cover it with two squares? Can you cover it with 58 squares? For how many different numbers of squares can you find a covering?”

You can, of course, explore this problem using graph paper. But for a digital experience, click on the link below and explore using Google Slides:

http://tinyurl.com/Squares100

When you click on that link, you’ll be required to “Make a copy” for your own Google Drive. This is done for two reasons: first, it’ll prohibit any editing to the original file, so that other folks who use that link will receive a similarly pristine copy; and second, it places a copy in your Google Drive that you can play with, modify, or share with your students. If you do that, be sure to change /edit and anything that follows at the end of the URL to /copy, which will require anyone you share it with to make a copy, too. Of course, you could also just share the link above with your students, and they can have a copy of the original file.

When using this problem in the classroom, start by showing a few examples — such as the ones above — and then have all students find a covering for the grid that’s different from your examples. Because every student can find at least one covering, this breeds confidence. In addition, it ensures that all students understand the problem. When all students have found a covering, have them share it with a partner. From then on, allow students to work in pairs or small groups to find other coverings. 

To facilitate the discussion, I draw a blank hundreds grid on the whiteboard. As unique coverings are found, students are allowed to enter the number in the hundreds grid. Of course, I require them to first verify with a partner that they have counted the number of squares correctly; once confirmed, they can show the covering to me; and if that number hasn’t been entered in the hundreds grid yet, they get to enter it. This can be motivating for students, and it’s a good opportunity to get less participatory students engaged.

When using this problem recently with a group of students, this is how the hundreds grid looked after about 20 minutes:

Whiteboard Record of Student Work

You’ll notice that numbers appear in different colors; each group received a different color dry erase marker. What you might also notice is that the numbers 17, 20, 23, 26, …, 65 all appear in red. This seemed more than coincidental, so I asked about it. The group members explained:

We realized we could cover the grid with a one 6 × 6 square and sixteen 2 × 2 squares:

We then divided one of the 2 × 2 squares into four 1 × 1 squares. That meant that one 2 × 2 square was replaced by four 1 × 1 squares, increasing the number of squares by three. So, the grid was now covered with 17 + 3 = 20 squares.

If we kept doing that — if we kept dividing the 2 × 2 squares — then we’d keep adding three more squares, so we could get 23, 26, 29, and so on, all the way up to 65.

The realization that one configuration could be transformed into many others allowed students to find coverings for myriad numbers. For what values of n could the grid not be covered? That’s left as a question for you. Have fun!

Two the Hard Way (and an Easy Way)

During our trip to Arizona for winter break, two problems surfaced organically while we were on holiday. (Sorry. Two math problems. There were lots of non-math problems, too, but we don’t have time for all that.)

On the plane, the interactive in-flight map showed the outside temp, toggling between Fahrenheit and Celsius. That led to the following MJ4MF original problem, which I thought — and still think — is pretty good:

The conversion between Fahrenheit and Celsius temperatures follows the rule F = 9/5 C + 32. Sometimes, the temperature is positive for both Celsius and Fahrenheit; sometimes, the temperature is negative for both Celsius and Fahrenheit; and other times, Fahrenheit is positive while Celsius is negative. What is the least possible product of the Fahrenheit temperature and its corresponding Celsius temperature?

You can pause here if you’d like to solve this before I present a spoiler.

Before I reveal two different solutions, allow me to digress. It could be that the statement about the temps sometimes being positive and sometimes being negative is denying a teachable moment. The graph below shows the linear relationship between the two temperature scales. Perhaps a good classroom question is:

When will the product CF be positive and when will it be negative?

Or maybe a better question is:

When are C and F both positive, when are they both negative, and when do they have different signs?

So, to the problem that I posed. As I thought about it on the plane, I concluded that if F = 1.8C + 32, then the product CF = 1.8C² + 32C. I then used calculus, found the derivative (CF)’ = 3.6C + 32, set that equal to 0, and concluded that C = ‑8.89, approximately. The corresponding Fahrenheit temperature is F = 16, so the minimum product is roughly ‑142.22.

Using calculus was like rolling a pie crust with a steamroller, though. I could have just as easily graphed the parabola and noted its vertex:

If I had used the other form of the rule, namely C = 5/9 (F ‑ 32), things might have been a little easier. Maybe. In that case, setting the derivative equal to 0 yields F = 16, which is arguably a nicer number. But then you still have to find the corresponding Celsius temperature, which is C = ‑8.89, and the product is still roughly ‑142.22. So, not much easier, if at all, and again graphing the parabola and noting its vertex would have done the trick:

The only real benefit to using this alternate version of the rule is that it provides a reasonable check. Since both methods — and both graphs — yield an answer of ‑142.22, we can feel confident in the result.

But there’s an easier way to solve this one.

Thinking this was a good problem — and because I like when my sons make me feel stupid — I gave it to Eli and Alex. Within seconds, Eli said, “Well, F is positive and C is negative between 0°F and 32°F, so the minimum will occur halfway between them at F = 16. That means C = ‑80/9, so it’s whatever ‑1280/9 reduces to.” (Turns out, -1280/9 = ‑142 2/9 ≈ ‑142.22.)

Eli hasn’t taken calculus, so he doesn’t know — or, at least, he hasn’t learned — that the minimum product should occur halfway between the x– and y‑intercepts of the linear graph. Yet, he had an intuitive insight that just happens to be true. As a result, what took me about five minutes of deriving and manipulating took him about five seconds.

The second problem arose at the grocery store. Among our purchases was a box of sugar cubes, which contained, surprisingly, 126 cubes.This number is surprising in the sense that it’s not a number you’ll see very often, except for an occasional appearance in the ninth row of Pascal’s Triangle, or maybe if you’re a chemist searching for stable atoms.

A question that could have arisen from this situation involves surface area and volume:

A rectangular prism with integer dimensions has a volume of 126 cubic units. What is the least possible surface area?

That’s not the question that was shared with Alex and Eli, though. (The answer, if you care, is 162 square units, which results from a 3 × 6 × 7 arrangement — which, in fact, is the exact arrangement of cubes in the box above. I suspect this is not a coincidence.)

The problem that I shared with my sons involved probability:

Imagine that the arrangement of cubes is removed from the box intact, and all six faces of the prism are painted red. Then one of the sugar cubes is selected at random and rolled. What is the probability that the top face of the rolled cube will be red?

The boys made an organized list, as follows:

Painted Faces	Number of Cubes
3	8
2	40
1	58
0	20

Further, the boys reasoned:

• P(cube with 3 red faces, red face lands on top) = 8/126 x 1/2 = 8/252
• P(cube with 2 red faces, red face lands on top) = 40/126 x 1/3 = 40/378
• P(cube with 1 red face, red face lands on top) = 58/126 x 1/6 = 58/756

Therefore,

• (P of getting a red face) = 8/252 + 40/378 + 58/576 = (24 + 80 + 58) / 576 = 162/576 = 9/42

Wow! That seems like a lot of work to get to the answer. Surely there’s an easier way, right?

Indeed, there is.

Notice that the penultimate step yielded the fraction 162/576. The numerator, 162, may look familiar. It’s the answer to the question that wasn’t asked above, the one about the least possible surface area of the prism. That’s no coincidence. In total, there will be 162 faces painted red. And there are 6 × 126 = 576 total faces on all of the sugar cubes (that is, six faces on each cube). This again suggests that the probability of rolling a red face is 162/576.

Did you happen to notice that the volume and surface area use the same digits in a different order? Cool.

So there you have it, two problems, each with two solutions, one easy and one hard. Or as mathematicians might say, one elegant and one common.

It’s typical for problems, especially problems worth solving, to have more than one solution strategy. What’s the trick to finding the elegant solution? Sadly, no such trick exists. Becoming a better problem solver is just like everything else in life; your skills improve with practice and experience. It’s akin to Peter Sagal’s advice in The Incomplete Book of Running, where he says, “You want to be a writer? […] Just sit down and write. The more you write, the better a writer you will become. You want to be a runner? Run when you can and where you can. Increase your mileage gradually, and your body will respond and you’ll find yourself running farther and faster than you ever thought possible.” You want to be a problem solver? Then spend your time solving problems. That’s the only way to increase the likelihood that you’ll occasionally stumble on an easy, elegant solution.

And every once in a while, you may even solve a problem faster than your kids.

12 Math Games, Puzzles, and Problems for Your Holiday Car Trip

It doesn’t matter if you’re one of the 102.1 million people traveling by car, one of the 6.7 million people traveling by plane, or one of the 3.7 million people traveling by train, bus, or cruise ship this holiday season — the following collection of games, puzzles, and problems will help to pass the time, and you’ll be there long before anyone asks, “Are we there yet?”

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Games

1. Street Sign Bingo

Use the numbers on street signs to create expressions with specific values. For instance, let’s say you see the following sign:

The two numbers on the sign are 12 and 3, from which you could make the following:

12 + 3 = 15

12 – 3 = 9

12 × 3 = 36

12 ÷ 3 = 4

Of course, you could just use the numbers directly as 3 or 12. Or if your passengers know some advanced math, they could use square roots, exponents, and more to create other values.

Can you split the digits within a number and use them separately? Can you concatenate two single-digit numbers to make a double-digit number? That’s for you and your traveling companions to decide.

To play as a competitive game, have each person in the car try to get every value from 1‑20. It’s easiest to keep track if you require that players go in order. To make it a cooperative game, have everyone in the car work collectively to make every value from 1 to 100.

As an alternative, you could use the numbers on license plates instead.

2. Bizz Buzz Bang

Yes, I’ve played this as a drinking game. No, I don’t condone drinking while driving. No, I don’t condone under-age drinking, either. Yes, I condone playing this game with minors while driving. (See what I did there?)

The idea is simple. You pick two single-digit numbers, A and B. Then you and your friends start counting, one number per person. But each time someone gets to a number that contains the digit A or is a multiple of A, she says, “Bizz!” Every time someone gets to a number that contains the digit B or is a multiple of B, she says, “Buzz!” And every time someone gets to a number that meets both criteria, he says, “Bang!”

For example, let’s say A = 2 and B = 3. Then the counting would go like this:

one, bizz, buzz, bizz, five, bang, seven, bizz, buzz, bizz, eleven, bang, buzz, …

When someone makes a mistake, that round ends. Start again, and see if you can beat your record.

You can use whatever numbers you like, or modify the rules in other ways. For instance, what if bizz is for prime numbers and buzz is for numbers of the form 4n + 3? Could be fun!

3. Dollar Nim (or any other variation)

Nim is a math strategy game in which players take turns removing coins from a pile. Different versions of the game are created by adjusting the number of coins that can be removed on each turn, the number of coins in the pile originally, how many piles there are, and whether you win (normal) or lose (misère) by taking the last coin.

A good aspect of Nim is that you don’t actually need coins. To play in the car, players just need to keep track of a running total in their heads.

A version of Nim dubbed 21 Flags was played on Survivor Thailand several years ago. There were 21 flags, and each team could remove 1, 2, or 3 flags on each turn. The team to remove the last flag won. This is a good first version to play in the car, especially for young kids who have never played before. Then mix it up by changing the initial amount and the number that can be removed on each turn.

Our family’s favorite version, Dollar Nim, is played by starting with $1.00 and removing the value of a common coin (quarter, dime, nickel, penny) on each turn. (While discussing this post with my sons, they informed me that they much prefer Euro Nim, which begins with 1€, but then the coin values to be removed are 50c, 20c, 10c, 5c, 2c, and 1c. It’s essentially the same game, but they like the different coin amounts.)

Three-Pile Nim consists of not just one but three piles with 3, 5, and 7 coins, respectively.  On each turn, a player must remove at least one coin, and may remove any number of coins, but all removed coins must be from the same pile.

Finally, Doubling Nim is played as the name implies. On the first turn, a player may remove any number of coins but not the entire pile. On every turn thereafter, a player may remove any number of coins up to double the number taken on the previous turn. For instance, if your opponent removes 7 coins, then you can remove up to 14 coins.

For every version of Nim, there is an optimal strategy. We’ve wasted hours on car trips discussing the strategy for just one variation. Discussing the strategies for all the versions above could occupy the entire drive from Paducah to Flint.

4. Guess My Number

One person picks a number, others ask questions to try to guess the number.

The simplest version is using “greater than” and “less than” questions. Is it greater than 50? Is it less than 175? And so forth. Using this method, the guessers can reduce the number of possibilities by half with each question, so at most, it should take no more than n guesses if 2ⁿ > m, where m is the maximum possible number that the picker may choose. For instance, if the picker is required to choose a number less than 100, then n = 7, because 2⁷ = 128 > 100. Truthfully, this version of the game gets boring quickly, but it’s worth playing once or twice, especially if there’s one picker and multiple guessers. And a conversation about the maximum number of guesses need can be a fun, mathy way to spend 15 minutes of your trip.

A more advanced version excludes “greater than” and “less than” questions. Instead, the guessers can ask other mathematical questions like, “Is it a prime number?” or “Do the digits of the number differ by 4?” Those questions imply, of course, that the answer must be yes or no, and that’s typical for these types of guessing games. If you remove that restriction, though, then guessers could ask questions that reveal a little more information, like, “What is the difference between the digits?” or “What is the remainder when the number is divided by 6?” With this variant, it’s often possible to identify the number with two strategic questions.

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Puzzles

Here are three puzzles that can lead to hours of conversation — and frustration! — on a car trip. Before you offer one to your crew, though, put forth the disclaimer that anyone who’s heard the puzzle before must remain mum. No reason they should spoil the fun for the rest of you. (The puzzles are presented here without solution, because you’ll know when you get the right answer. You can find the answer to any of them online with a quick search… but don’t do that. You’ll feel much better if you solve it yourself.)

5. Dangerous Crossing

Four people come to a river in the middle of the night. There’s a narrow bridge, but it’s old and rickety and can only hold two people at a time. They have just one flashlight and, because it’s night and the bridge is in disrepair, the flashlight must be used when crossing the bridge. Aakash can cross the bridge in 1 minute, Britney in 2 minutes, Cedric in 5 minutes, and Deng in 8 minutes. When two people cross the bridge together, they must travel at the slower person’s pace. And they need to hurry, because zombies are approaching. (Oh, sorry, had I failed to mention the zombie apocalypse?) What is the least amount of time that all four people can cross the bridge?

And how can you be sure that your method is the fastest?

6. Weight of Weights

Marilyn has a simple balance scale and four small weights, each weighing a whole number of grams. With the balance scale and these weights, she is able to determine the weight of any object that weighs between 1 kg and 40 kg. How much does each of the four weights weigh?

7. Product Values

Assign each letter a value equal to its position in the alphabet, i.e., A = 1, B = 2, C = 3, …, Z = 26. Then for any common word, find its product value by multiplying the value of the letters in the word. For instance, the product value of CAT is 60, because 3 × 1 × 20 = 60.

1. Find as many common English words as you can with a product value of 60.
2. Find a common English word that has the same product value as your name. (A little tougher.)
3. Find a common English word with a product value of 3,000,000. (Zoiks!)

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Problems

8. The Three of Life

This one looks so innocent!

What’s the probability that a randomly chosen number will contain the digit 3?

But spend a little time with it. And prepare for. Mind. Blown.

9. Hip to Be (Almost) Square

No calculators for this one.

There are four positive numbers — 1, 3, 8, and x — such that the product of any two of them is one less than a square number. What is the least possible value of x?

No spreadsheets, either.

10. Hip to Be Square (Roots)

Just some good, old-fashioned algebra and logic to tackle this one.

Which of the following expressions has a greater value?

  or 

Surprised?

11. Coming and Going

Potentially counterintuitive.

At the holidays, Leo drove to his grandma’s, and the traffic was awful! His average speed was 42 miles per hour. After the holidays, however, he drove home along the same route, and his average speed was 56 miles per hour. What was his average speed for the entire trip?

And, no, your first guess was most likely not correct.

12. The Year in Numbers

A moldy oldie, to be sure, but this puzzle is always a crowd-pleaser for those who haven’t seen it before.

Use the digits of the new year — 2, 0, 1, and 9 — and any mathematical operations to form the integers from 1 to 100. For instance, you can form 1 as follows: 2 × 0 × 9 + 1 = 1.

For an added challenge, add the restriction that you must use the four digits in order.

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Bonus: A Book

My sons get sick when they read in the car. That’s why I love the two books Without Words and More Without Words by James Tanton. Literally, there are no words! Each puzzle is presented using a few examples, and then students must follow the same rules to solve a few similar, but more challenging, puzzles.

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Wherever you’re headed during the holiday break — driving to your relatives’ house, flying to Fort Lauderdale, or just relaxing at home — I hope your holidays are filled with joy, happiness, and lots of math!

Stick Figure Math

I’ll never forget the first time I saw the pattern

1, 2, 4, 8, 16, __

and was dumbfounded to learn that the missing value was 31, not 32, because the pattern was not meant to represent the powers of 2, but rather, the number of pieces into which a circle is divided if n points on its circumference are joined by chords. Known as Moser’s circle problem, it represents the inherent danger in making assumptions from a limited set of data.

Last night, my sons told me about the following problem, which they encountered on a recent math competition:

What number should replace the question mark?

Well, what say you? What number do you think should appear in the middle stick figure’s head?

Hold on, let me give you a hint. This problem appeared on a multiple-choice test, and these were the answer choices:

1. 3
2. 6
3. 9
4. 12

Now that you know one of those four numbers is supposed to be correct, does that change your answer? If you thought about it in the same way that the test designers intended it, then seeing the choices probably didn’t change your answer. But if you didn’t think about it that way and you put a little more effort into it, and you came up with something a bit more complicated — like I did — well, then, the answer choices may have thrown you for a loop, too, and made you slap your head and say, “WTF?”

For me, it was Moser’s circle problem all over again.

So, here’s where I need your help: I’d like to identify various patterns that could make any of those answers seem reasonable.

In addition, I’d also love to find a few other patterns that could make some answers other than the four given choices seem reasonable.

For instance, if the numbers in the limbs are a, b, c, and d, like this…

then the formula 8a – 4d gives 8 for the first and third figures’ heads and yields 8 × 6 – 4 × 9 = 12 as the answer, which happens to be one of the four answer choices.

Oh, wait… you’d don’t like that I didn’t use all four variables? Okay, that’s fair. So how about this instead: ‑3a + b + c + 2d, which also gives ‑3 × 6 + 7 + 5 + 2 × 9 = 12.

Willing to help? Post your pattern(s) in the comments.

[UPDATE (3/9/18): I sent a note to the contest organizers about this problem, and I got the following response this afternoon: “Thanks for your overall evaluation comments on [our] problems, and specifically for your input on the Stick Figure Problem. After careful consideration, we decided to give credit to every student for this question. Therefore, scores will be adjusted automatically.”]

The Homework Inequality: 1 Great Problem > 50 Repetitive Exercises

Yesterday, my sons Alex and Eli were completing their homework on fraction operations, which included 39 problems in 4 sets, wherein each problem in a set was indistinguishable from its neighbors.

The worksheet contained 12 problems of fraction addition, 9 problems of fraction subtraction, 9 problems of fraction multiplication, and 9 problems of fraction-times-whole-number multiplication. That’s 39 problems of drudgery, when 10 problems would’ve been sufficient. Here’s a link to the worksheet they were given, if you’d like to torture your children or students in a similar way:

Lesson 0.1: Adding and Multiplying Fractions

Frustrated by the monotony of the assignment, I told the boys they didn’t have to do all of the problems, and they could stop when they felt that they had done enough from each set.

“No,” said Alex. “We’re supposed to do them all.”

My sons are responsible students, but I’m frustrated by teachers who take advantage of their work ethic. Just because they’re willing to complete 50 exercises for homework doesn’t mean they should be assigned 50 exercises for homework.

My colleague at Discovery Education, Matt Cwalina, puts it this way:

Some say that a picture is worth a thousand words. I say,

A great problem is worth a thousand exercises.

Personally, I would much rather have students think deeply about one challenging problem than mindlessly complete an entire worksheet. Luckily, my sons take after their daddy and love number puzzles, so I spontaneously created one.

Find three fractions, each with a single-digit numerator and denominator, that multiply to get as close to 1 as possible. Don’t repeat digits.

Eli started randomly suggesting products. “What about 4/5 × 6/7 × 9/8?” He’d work out the result, say, “I think I can do better,” then try another. And another. And another. Finally, he found a product that equaled 1. (No spoiler here. Find it yourself.)

Alex eventually found an answer, too. At the bottom of his homework assignment, he added a section that he titled “Bonus” where he captured his attempts:

I don’t know exactly how many calculations Eli completed while working on this problem, but I know that Alex completed at least seven, thanks to his documentation. Wouldn’t you agree that completing several fraction computations while thinking about this more interesting problem is superior to doing a collection of random fraction computations with no purpose?

There is a preponderance of evidence (see Rohrer, Dedrick, and Stershic 2015; definitely check out Figure 4 at the top of page 905) that massed practice — that is, completing a large number of repetitions of the same activity over and over — is counterproductive. Unfortunately, massed practice feels good because it results in short-term memory gains, which trigger a perceived level of mastery; but, it doesn’t lead to long-term retention. Moreover, students who learn a skill by practicing it repeatedly get really good at performing that skill when they know it’s coming; but, two months down the road, when they need to use that skill in an unfamiliar context because it’s not on a worksheet titled “Lesson 0.1: Adding and Multiplying Fractions,” they’re less likely to remember than if they had used more effective practice methods. One of those more effective methods is interleaving, which involves spacing out practice over multiple sessions and varying the difficulty of the tasks. Whether you’re trying to learn how to integrate by parts or how to hit a curve ball, be sure to make your practice exercises a little more difficult than you’re used to. Know that interleaving your practice will not feel as good as massed practice while you’re doing it; but later, you’ll feel better due to improved memory, long-term learning, and mastery of skills.

Interleaving is one of the reasons I love the MathCounts School Handbook, which can be downloaded for free from the MathCounts website. The topics covered by the 250 problems in the School Handbook run the gamut from algebra, number sense, and probability, to geometry, statistics, combinatorics, sequences, and proportional reasoning — and any given page may contain problems of any type! Veteran MathCounts coach Nick Diaz refers to this mixture as “shotgun style,” meaning that students never know what’s coming next. Consequently, similar problems are not presented all at once; instead, students are exposed to them several days or perhaps weeks apart. Having to recall a skill that hasn’t been used for a while requires more effort than remembering what you did just five minutes ago, but the result is long-term retention. It’s doubtful that the writers of the first MathCounts School Handbook knew the research about interleaving and massed practice… but they clearly knew about effective learning.

The other reason I like the MathCounts School Handbook is the difficulty level of the problems. Sure, some of the items look like traditional textbook exercises, but you’ll also find a lot of atypical problems, like this one from the 2017-18 School Handbook:

If p, q, and r are prime numbers such that pq + r = 73, what is the least possible value of p + q + r?

That problem, as well as the fraction problem that I created for Alex and Eli, would both fall into the category of open-middle problems, which means…

• the beginning is closed: every students starts with the same initial problem.
• the end is closed: there is a small, finite number of unique answers (often, just one).
• the middle is open: there are multiple ways to approach and ultimately solve the problem.

Open-middle problems often allow for implicit procedural practice while asking students to focus on a more challenging problem. This results in a higher level of engagement for students. Moreover, it reduces the need for massed practice, because students are performing calculations while doing something else. You can find a large collection of open-middle problems at www.openmiddle.com, and the following is one of my favorites:

Use the digits 1 to 9, at most one time each, to fill in the boxes to make a result that has the greatest value possible.

It’s a great problem, because random guessing will lead students to combinations that work, but it may not be obvious how to determine the greatest possible value. Consequently, there’s an entry point for all students, the problem offers implicit procedural practice, and the challenge of finding the greatest value provides motivation for students to continue.

I have a dream that one day, in traditional classrooms where 50 problems are assigned for homework every night, where procedural fluency is valued over conceptual understanding; that one day, right there in those classrooms, students will no longer think that math is simply a series of disparate rules with no purpose, but instead will experience the joy of attempting and solving challenging problems that inspire purposeful play and, as a side benefit, encourage students to practice the skills they will need to be successful learners.

There are myriad resources available so that teachers and parents can encourage their students to engage in these kinds of problem-solving activities, so it is my hope that this dream is not too far away.

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As a special bonus for reading to the end, check out this Interleaved Mathematics Practice Guide that Professor Doug Rohrer was kind enough to share with me (and now, with you).

Σ Π :: The Sum and Product Game

This joke, or a close facsimile, has been taking a tour of email servers recently, and it’s now showing up on t-shirts, too:

…and it was delicious!

Appropriate for Pi Day, I suppose, as is the game my sons have been playing…

Eli said to Alex, “18 and 126.”

Alex thought for a second, then replied, “2, 7, and 9.”

“Yes!” Eli exclaimed.

I was confused. “What are you guys doing?” I asked.

“We invented a game,” Eli said. “We give each other the sum and product of three numbers, and the other person has to figure out what the numbers are.”

After further inquisition, I learned that it wasn’t just any three numbers but positive integers only, that none can be larger than 15, and that they must be distinct.

Hearing about this game made me immediately think about the famous Ages of Three Children problem:

A woman asks her neighbor the ages of his three children.

“Well,” he says, “the product of their ages is 72.”

“That’s not enough information,” the woman replies.

“The sum of their ages is your house number,” he explains further.

“I still don’t know,” she says.

“I’m sorry,” says the man. “I can’t stay and talk any longer. My eldest child is sick in bed.” He turns to leave.

“Now I know how old they are,” she says.

What are the ages of his children?

You should be able to solve that one on your own. But if you’re not so inclined, you can resort to Wikipedia.

But back to Alex and Eli’s game. It immediately occurred to me that there would likely be some ordered pairs of (sum, product) that wouldn’t correspond to a unique set of numbers. Upon inspection, I found eight of them:

(19, 144)
(20, 90)
(21, 168)
(21, 240)
(23, 360)
(25, 360)
(28, 630)
(30, 840)

My two favorite ordered pairs were:

(24, 240)
(26, 286)

I particularly like the latter one. If you think about it the right way (divisibility rules, anyone?), you’ll solve it in milliseconds.

And the Excel spreadsheet that I created to analyze this game led me to the following problem:

Three distinct positive integers, each less than or equal to 15, are selected at random. What is the most likely product?

Creating that problem was rather satisfying. It was only through looking at the spreadsheet that I would’ve even thought to ask the question. But once I did, I realized that solving it isn’t that tough — there are some likely culprits to be considered, many of which can be eliminated quickly. (The solution is left as an exercise for the reader.)

So, yeah. These are the things that happen in our geeky household. Sure, we bake cookies, play board games, and watch cartoons, but we also listen to the NPR Sunday Puzzle and create math games. You got a problem with that?

Math Problem for 2017

Happy New Year! Welcome to 2017.

Here are some interesting facts about the number 2017:

• It’s prime. (Okay, so that fact isn’t very interesting. But just you wait…)
• Insert a 7 between any two digits of 2017, and the result  is still a prime number. That is, 27,017, 20,717, and 20,177 are all prime. (See? Told you it was gonna get better.)
• The cube root of 2017 is approximately 12.63480759, which uses all ten digits 0‑9, and 2017 is the least positive integer that has this property. (Mind blown yet?)
• The decimal expansion of 2017²⁰¹⁷ has 6,666 digits.
• 2017 = 44² + 9²
• 2017 = 12³ + 6³ + 4³ + 2³ + 1³ = 10³ + 9³ + 6³ + 4³ + 2³

If you need some more, check out Matt Parker’s video.

Sorry, no video from me. But in honor of our newly minted prime year, I have created a problem for you to solve.

In the area model below (not to scale), the area of the five blue regions is indicated by the number inside the rectangle. What is the area of the yellow region with the question mark inside?

Sorry, I don’t give answers. Feel free to have at it in the comments.