## Posts tagged ‘problem’

### 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:

$\sqrt{-1} \hspace{0.2 in} 2^3 \hspace{0.2 in} \Sigma \hspace{0.2 in} \pi$
…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.

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.

### 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.

• 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 frustrated.

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 25n + 1 to 25n + 25, where n is the number of quarters to be returned, you’ll receive a nickel when the amount of change is 25n + k, where 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 176,432,928 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.
• 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.
• 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 2n if the last n digits are divisible by 2n.

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, and d are four distinct positive integers such that ab = cd, what is the least possible value of 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.

### 16 Math Problems for 2016

Yes, I know that I just posted some Math Problems for 2016 on December 19.

But I’ve decided to post some more for a variety of reasons:

• 2016 is cool.
• It’s a triangular number.
• It has lots of factors. (I’d tell you exactly how many, except that’s one of the problems below.)
• After writing the problems for that previous post, I just couldn’t control myself.
• It’s my blog, and I can do what I want.

People who write math problems for competitions (like me) love to be cheeky and include the year number in a problem, especially when any sufficiently large number will do. When the year number is critical to the success of a problem, well, that’s just a bonus. With that in mind, there are 16 problems below, each of which includes the number 2016.

A fully formatted version of these problems, complete with answer key, extensions, and solutions, is available for purchase through the link below:

16 Problems for 2016 — just \$1

Enjoy, and happy new year!

1. What is the sum of 2 + 4 + 6 + 8 + ··· + 2016?
1. Using only common mathematical symbols and the digits 2, 0, 1, and 6, make an expression that is exactly equal to 100.
1. Find a fraction with the following decimal equivalent.

$0.\overline{2016}$

1. How many positive integer factors does 2016 have?
1. What is the value of n if 1 + 2 + 3 + ··· + n = 2016?
1. Find 16 consecutive odd numbers that add up to 2016.
1. Create a 4 × 4 magic square in which the sum of each row, column, and diagonal is 2016.
1. Find a string of two or more consecutive integers for which the sum is 2016. How many such strings exist?
1. What is the value of the following series?

$\frac{1}{1} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{5} + \cdots + \frac{1}{2015} \times \frac{1}{2016}$

1. What is the units digit of 22016?
1. Some people attend a party, and everyone shakes everyone else’s hand. A total of 2016 handshakes occurred. How many people were at the party?
1. What is the value of the following expression, if x + 1/x = 2?

$x^\mathbf{2016} + \frac{1}{x^\mathbf{2016}} + \mathbf{2016}$

1. A number of distinct points were placed along the circumference of a circle. Each point was then connected to every other point, and a total of 2016 segments were formed. How many points were placed on the circle?
1. Let A = 1, B = 2, C = 3, …, Z = 26. Find a word for which the product of the letters is 2016. (This one may look familiar.)
1. Each dimension of a rectangular box is an integer number of inches. The volume of the box is 2016 in3. What is the least possible surface area of the box?
1. What is the maximum possible product for a set of positive integers that have a sum of 2016?

UPDATE: Bonus Material!

Special thanks to my friend Harold Reiter, who created the following 2016 problems for use as MathCounts practice:

• What is the smallest number N such that the product of the digits of N is 2016?
• What is the sum of the divisors of 2016?
• What is the product of the divisors of 2016? Express your answer as a product of prime numbers.
• Solve the following equation:

$\binom{n}{2} = 2016$

• What is the binary representation of 2016?
• What is the base-4 representation of 2016?
• What is the base-8 representation of 2016?

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

## MJ4MF (offline version)

Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.