## Posts tagged ‘problem’

### 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= 9/5C+ 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

CFbe positive and when will it be negative?

Or maybe a better question is:

When are

CandFboth 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* = 1.8*C* + 32, then the product *CF* = 1.8*C*^{2} + 32*C*. I then used calculus, found the derivative (*CF*)’ = 3.6*C* + 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?”

**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 =

1512 – 3 =

912 × 3 =

3612 ÷ 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 4*n* + 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 2^{n} > *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 2^{7} = 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.

**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.

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

**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 ofx?

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*.

**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.

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:

- 3
- 6
- 9
- 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 8*a* – 4*d* 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: ‑3*a* + *b* + *c* – 2*d*, 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:

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, andrare prime numbers such thatpq+r= 73, what is the least possible value ofp+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.

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 2017
^{2017}has 6,666 digits. - 2017 = 44
^{2}+ 9^{2} - 2017 = 12
^{3}+ 6^{3}+ 4^{3}+ 2^{3}+ 1^{3}= 10^{3}+ 9^{3}+ 6^{3}+ 4^{3}+ 2^{3}

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.

### It’s Not What’s on the Outside…

Through the Academic and Creative Endeavors (ACE) program at their school, my sons participate in the Math Olympiad for Elementary and Middle School (MOEMS). While passing the door to the ACE room yesterday, I noticed a sign with the names of those who scored a perfect 5 out of 5 on the most recent contest — and my sons’ names were conspicuously absent. Last night at dinner, I asked Alex and Eli what happened, and they told me about the problem that they both missed. (What? Like we’re the only family in America that discusses math problems at the dinner table.) Here’s how they explained it to me:

Nine 1-cm by 1-cm squares are arranged to form a 3 × 3 square, as shown below. The 3 × 3 square is divided into two pieces by cutting along gridlines only. What is the greatest total (combined) perimeter for the two shapes?

*The answer to this problem appears below. Pause here if you’d like to solve it before reaching the spoiler.*

[Ed. Note: I didn’t see the actual exam, so the presentation of the problem above is based entirely on my sons’ description. Apologies to MOEMS for any substantive differences.]

This problem epitomizes what I love about math competitions.

**The answer to the problem is not obvious.**This is the case with many competition problems, unlike the majority of problems that appear in a traditional textbook.**The solution does not rely on rote mechanics.**Again, this differentiates it from a standard textbook problem or — shudder! — from the problems that often populate the databases of many skill-based online programs.**Students have to get messy.**That is, they’ll need to try something, see what happens, then decide if they can improve the result.**Students have to convince themselves when to stop.**Or more precisely, they’ll need to convince themselves that they’ve found the correct answer. For instance, let’s say a student divides the square into a 3 × 2 and a 3 × 1 rectangle. The combined perimeter is 18 units. Is that good enough, or can you do better? This is different from, say, a typical algebra problem, for which students are taught how to check their answer.

The problem also epitomizes what many people hate about math competitions.

**There’s a time limit.**Students have 26 minutes to solve 5 problems. Which means that if students spend more than 5 minutes on this one, they may not have time to finish the other four. (There was a student of John Benson who, when asked about his goal for an upcoming math competition, replied, “I hope to solve half the problems during the competition and all of them by the end of the week.” That’s the way mathematicians work.)**It’s naked math.**Sorry, nothing real-world about this one. (But maybe that’s okay, because real may not be better.)**The problem is presented as a neat little bundle.**This is rarely how mathematics actually works. True problems often don’t present themselves all at once; it’s through investigation and research that the constraints become known and the nuances are revealed.

All that said, *I believe that the pros far outweigh the cons*. Benjamin Franklin Finkel said, “Many dormant minds have been aroused into activity through the mastery of a single problem.” I don’t remember the last time a mind was aroused by the solution to *x* + 7 = -3, but I’ve witnessed awakenings when students solve problems like the one above.

And here’s the tragedy in all of this: **Many teachers believe that only kids who participate in math competitions can handle — or appreciate — math competition questions.** No! Quite the opposite, in fact. Students who have tuned out have done so because they’ve never been challenged and, worse, have never felt the thrill of solving a problem on their own.

What I really love about the problem, though, is it made me think about other questions that could be asked:

- How many ways are there to divide a 3 × 3 square into two pieces that will yield the maximum total perimeter?
- What is the maximum total perimeter if a 3 × 3 square is divided into three pieces?
- What is the maximum total perimeter if a 4 × 4 square is divided into two pieces? …a 5 × 5 square? …a 6 × 6 square?
*The answer to the problem about the 4 × 4 square appears below. Pause here if you’d like to solve it before reaching the spoiler.*

- (wait for it) What is the maximum total perimeter if an
*n*×*n*square is divided into two pieces?

It was that last question that really got the blood pumping.

Here’s a solution for how to divide a 3 × 3 square to yield the greatest total perimeter:

And here’s a solution for how to divide a 4 × 4 square to yield the greatest total perimeter:

What’s the solution for an *n* × *n* square? That’s left as an exercise for the reader.

### Dos Equis XX Math Puzzles

No, the title of this post does not refer to the beer. Though it may be the most interesting blog post in the world.

It refers to the date, 10/10, which — at least this year — is the second day of National Metric Week. It would also be written in Roman numerals as X/X, hence the title of this post.

For today, I have not one, not *two*, but **three** puzzles for you. I’m providing them to you well in advance of October 10, though, in case you’re one of those clever types who wants to use these puzzles on the actual date… this will give you time to plan.

The first is a garden-variety math problem based on the date (including the year).

Today is 10/10/16. What is the area of a triangle whose three sides measure 10 cm, 10 cm, and 16 cm?

Hint:A triangle appearing in an analogous problem exactly four years ago would have had the same area.

The next two puzzles may be a little more fun for the less mathy among us — though I’m not sure that any such people read this blog.

Create a list of words, the first with 2 letters, the second with 3 letters, and so on, continuing as long as you can, where each word ends with the letter X. Scoring is triangular: Add the number of letters in all the words that you create until your first omission. For instance, if you got words with 2, 3, 4, 5, and 8 letters, then your score would be 2 + 3 + 4 + 5 = 14; you wouldn’t get credit for the 8-letter word since you hadn’t found any 6- or 7-letter words.

2 letters: _________________________

3 letters: _________________________

4 letters: _________________________

5 letters: _________________________

6 letters: _________________________

7 letters: _________________________

8 letters: _________________________

9 letters: _________________________

10 letters: _________________________

11 letters: _________________________

12 letters: _________________________

13 letters: _________________________

14 letters: _________________________

Note: There are answer blanks above for words up to 14 letters, because — you guessed it — the longest English word that ends with an X contains 14 letters.

The third and final puzzle is a variation on the second.

How many words can you think of that contain the letter X

twice? (Zoiks!) Scoring: Ten points for the first one, and a bazillion points for each one thereafter — this is hard! Good luck!

If you’re in desperate need of help, you can access **my list of words for both puzzles** — of which I’m fairly proud, since my list of words that end in X include math words for 2 through 10 letters — or do a search at www.morewords.com.

### Constant Change

I’m also old, cranky, and cynical. Whatever.

My frustration is not the my-flight-was-delayed-three-times-then-eventually-cancelled-and-there-are-no-more-flights-to-Cleveland-till-tomorrow-morning type. It’s not even the can’t-believe-my-boss-is-making-me-go-to-Cleveland kind of frustration. More like the why-aren’t-there-the-same-number-of-hot-dogs-and-buns-in-a-pack variety. So it’s a First World problem, to be sure, but still annoying. I’ll explain more in a moment.

But first, how ’bout a math problem to get us started?

If you make a purchase and pay with cash, what’s the probability that you’ll receive a nickel as part of your change?

Sure, if you want to get all crazy about this, then we can take all the fun out of this problem by stating the following assumptions:

- You only pay with paper currency. If you paid with coins, then the distribution of coins you’d receive as change would likely vary quite a bit.
- You never use 50¢ coins. Honestly, they’re just too obscure.
- Transaction amounts are uniformly distributed, so that you’re just as likely to receive 21¢ as 78¢ or any other amount.
- Cashiers don’t round because they dislike pennies. So, if you’re supposed to get 99¢ change, the cashier doesn’t hand you a dollar and say, “Don’t worry about it.” Instead, you actually get 99¢ change.

But stating assumptions is a form of mathematical douchebaggery, isn’t it? (As an aside, check out the definition of douchey that’s returned when you do a search. Sexist, anyone?) I prefer problems with no assumptions stated; let folks make their own assumptions to devise a model. If you and I get different answers because of different assumptions, no worries. Maybe we both learn something in the process.

Anyway, where was I? Oh, yeah…

Understanding the solution to that problem is a precursor to the issue that’s causing me frustration. I’ll give the solution in a minute, so pause here if you want to solve it on your own, but let me now allow the proverbial cat out of its bag and tell you why I’m frustrated.

At our local grocery store, there’s a **coin counting machine** that will count your change, sort it, and spit out a receipt that you can take to the customer service desk to exchange for paper currency. Walk in with a jar full of change, walk out with a fistful of fifties. Pretty nifty, right? Except the machine charges a ridiculous 8.9% fee to perform this service. No, thank you.

My bank used to have a similar coin counting machine, and if you deposited the amount counted by the machine into your account, there was no fee. The problem is that everyone was doing this to avoid the grocery store fee, so the machine broke often. The bank finally decided the machine wasn’t worth the maintenance fees and got rid of it. Strike two.

Which brings me to my current dilemma. One Saturday morning every month, we now spend 30 minutes counting coins and allocating them to appropriate wrappers. Which is fine. The problem, however, is that we run out of quarter and penny wrappers way faster than we run out of nickel or dime wrappers. Which brings me to the real question for the day:

Since pennies, nickels, dimes, and quarters are not uniformly distributed as change,

why the hell does every package of coin wrappers contain the same number for each coin type?

The Royal Sovereign Assorted Coin Preformed Wrappers is the best-selling collection of coin wrappers on Amazon, and it provides 54 wrappers for each coin type. They also offer a 360‑pack with 90 wrappers for each coin type; Minitube offers a 100‑pack with 25 wrappers for each coin type; and Coin-Tainer offers a 36‑pack with 9 wrappers for each coin type. But what **no one** offers, so far as I can tell, is **a collection of coin wrappers with a distribution that more closely resembles the distribution of coins that are received as change**.

Whew! It feels good to finally raise this issue for public consideration.

So, the question that I really wanted to ask you…

Given the distribution of quarters, dimes, nickels, and pennies that are received in change, and given the number of coins needed to fill a coin wrapper — 40 quarters, 50 dimes, 40 nickels, and 50 pennies — how many of each wrapper should be sold in a bundled collection?

To answer this question, I determined the number of coins of each type required for every amount of change from 1¢ to 99¢. The totals yield the following graph:

The number of pennies is nearly five times the number of nickels. And there are nearly twice as many quarters as dimes.

But I realize that’s a **theoretical** result that may not match what happens in practice, since this assumes that the amounts of change from 1¢ to 99¢ are uniformly distributed (they aren’t) and that cashiers don’t round down to avoid dealing with pennies (they do). In fact, when I made a purchase of $2.59 yesterday, instead of getting one penny, one nickel, one dime, and one quarter as change, the cashier gave me one penny, **three** nickels, and one quarter, in what was clearly a blatant attempt to skew my data.

So for an **experimental** result, I counted the pennies, nickels, dimes, and quarters in our home change jar. The results were similar:

The ratio of pennies to nickels is closer to three, but the ratio of quarters to dimes is still roughly two.

Using a hybrid of the theoretical and experimental results, and accounting for the fact that only 40 quarters and nickels are needed to fill a wrapper whereas 50 pennies and dimes are needed, it seems that an appropriate ratio of coin wrappers would be:

**quarters : dimes : nickels : pennies :: 17 : 8 : 6 : 19**

Okay, admittedly, that’s a weird ratio. Maybe something like **3:2:1:4**, to keep it simple. Or even 2:1:1:2. All I know is that 1:1:1:1 is completely insane, and this nonsense has got to stop.

Hello, Royal Sovereign, Minitube, and Coin-Tainer? Are you listening? I’ve completed this analysis for you, free of charge. Now **do the right thing**, and adjust the ratio of coin wrappers in a package accordingly. Thank you.

Wow, that was a long rant. Sorry. If you’ve made it this far, you deserve some comic relief.

How many mathematicians does it take to change a light bulb?

Just one. She gives it to a physicist, thus reducing it to a previously solved problem.If you do not change direction, you may end up where you are heading. – Lao Tzu

The only thing that is constant is change. – Heraclitus

Turn and face the strange ch-ch-ch-changes. – David Bowie

A Buddhist monk walks into a Zen pizza parlor and says, “Make me one with everything.” The owner obliged, and when the pizza was delivered, the monk paid with a $20 bill. The owner put the money in his pocket and began to walk away. “Hey, where’s my change?” asked the monk. “Sorry,” said the owner, “change must come from within.”

As for the “probability of a nickel” problem that started this post, here’s my solution.

For change amounts from 1¢ to 25¢, there are ten values (5‑9 and 15‑19) for which you’ll receive a nickel as part of your change.

This pattern then repeats, such that for change amounts from **25 n + 1** to

**25**, where

*n*+ 25*n*is the number of quarters to be returned, you’ll receive a nickel when the amount of change is

**25**, where

*n*+*k**k*∈ {5, 6, 7, 8, 9, 15, 16, 17, 18, 19}. For 0 ≤

*n*< 4, there are 40 different amounts of change that will contain a nickel, so the probability of getting a nickel as part of your change is 40/100, or 40%.

### Problems at the 2016 MathCounts National Competition

Yesterday, **Edward Wan** (WA) became the 2016 MathCounts National Champion. He defeated Luke Robitaille (TX) in the finals of the Countdown Round, 4-3. In the Countdown Round, questions are presented one at a time, and the first student to answer four correctly claims the title.

“This one is officially a nail-biter,” declared Lou DiGioia, MathCounts Executive Director and the moderator of the Countdown Round. Three times, Wan took a one-question lead; and three times, Robitaille tied the score on the following question. The tie was broken for good when Wan answered the following question:

What is the remainder when 999,999,999 is divided by 32?

This year’s winning question was relatively easy. What makes me say that? Well, for starters, when an odd number is divided by an even number, the remainder will be odd; and because 32 is the divisor, the remainder has to be less than 32. Consequently, the remainder is in the set {1, 3, 5, …, 31}, so there are only 16 possible answers.

But more importantly, most MathCounts competitors will be well trained for a problem of this type. It relies on divisibility rules that they should know, and it requires minimal insight to arrive at the correct answer.

I suspect that the following explanation of the solution is the likely thought process that Wan used to solve this problem; of course, all of this occurred **in his head in less than 7 seconds**, which does make it rather impressive.

A fact that you probably know:

- A number is divisible by 2 if it’s even.

But said another way…

- A number is divisible by 2 if the last digit is divisible by 2.

There are then corollary rules for larger powers of 2:

- A number is divisible by 4 if the last
**two**digits are divisible by 4.- For example, we can conclude that 176,432,9
**28**is divisible by 4 because the last two digits form 28, which is divisible by 4. The digits in the hundreds, thousands, and higher place values are somewhat irrelevant, because they represent some multiple of 100 — for instance, the 7 in the ten millions place represents 70,000,000, which is 700,000 × 100 — and every multiple of 100 is divisible by 4.

- For example, we can conclude that 176,432,9
- A number is divisible by 8 if the last
**three**digits are divisible by 8.- For example, we can conclude that 176,432,
**376**is divisible by 8 because the last three digits form 376, which is divisible by 8 since 8 × 47 = 376.

- For example, we can conclude that 176,432,
- A number is divisible by 16 if the last
**four**digits are divisible by 16. - A number is divisible by 32 if the last
**five**digits are divisible by 32. - And so on.

These observations lead to a generalization…

- A number is divisible by 2
if the last^{n}*n*digits are divisible by 2.^{n}

I won’t take the time to prove that statement here, but you can trust me. (Or maybe you’d like to prove it on your own.) I will, however, explain why it’s relevant.

A number will be divisible by 32 if the last five digits are divisible by 32. Consequently, any number that ends in five 0’s will be divisible by 32, which means that 1,000,000,000 is a multiple of 32. Since 999,999,999 is 1 less than 1,000,000,000, then it must be 1 less than a multiple of 32. Therefore, when 999,999,999 is divided by 32, **the remainder will be 31**.

The hardest part of solving that problem is recognizing that 999,999,999 is 1 less than a multiple of 32. But for most MathCounts students, that step is not very difficult, hence my contention that this was a relatively easy winning problem.

My favorite problem of the Countdown Round? Now, that’s another story, and it epitomizes what I generally love about MathCounts problems.

If

a,b,c, anddare four distinct positive integers such thata=^{b}c, what is the least possible value of^{d}a+b+c+d?

This problem has several things going for it:

- It’s simply stated.
- It’s easily understood, even by students who don’t participate in MathCounts.
- It has an entry point for all students, since most kids can find at least one set of numbers that would work, even if they couldn’t find the set with the least possible sum.
- Finding the right answer requires convincing yourself that no lesser sum exists.

It’s that last point that I find so interesting. While I was able to find the correct answer, it took a while to convince myself that it was the least possible sum. But since I don’t want to deprive you of any fun, I’ll let you solve the problem on your own.

As a final point, I’ll show you a picture that I took at the event. Do you see the error? What can I say… it’s a math competition… you didn’t expect them to be good with numbers, did you?

*Full disclosure: The error was corrected halfway through the competition during a break.*