Posts tagged ‘problem’

Constant Change

Coin PileI’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?

Coin WrappersThe 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:

Change - Theoretical

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:

Change - Experimental

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%.

August 30, 2016 at 5:10 am 3 comments

Problems at the 2016 MathCounts National Competition

Yesterday, EdwaEdward Wanrd 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.

MathCounts Logo

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?

MathCounts 2015 2016

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

May 10, 2016 at 4:09 pm 3 comments

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?

January 1, 2016 at 2:16 am 3 comments

Math Problems for 2016

“What homework do you have to do tonight?”

I ask my sons this question daily, when I’m trying to determine if they’ll need to spend the evening doing word study or completing a math worksheet, or if we’ll instead be able to waste our time watching The Muppets or, perhaps, pulling up the animated version of Bob and Doug Mackenzie’s 12 Days of Christmas on Dailymotion.

When I asked this question last night, though, the answer was surprising:

We have to do our reading, but we already completed your math problem.

My problem? I had no idea what this meant. So they explained:

It’s not a problem you gave us. It’s one we got from [our teacher], and it says, “This problem was written by Patrick Vennebush.”

I was puzzled, but then it dawned on me. I asked, “Does it have a monkey at the top with the word BrainTEASERS?”

“Yes!”

“Which problem?”

“It’s about the word CAT.”

I knew the problem immediately. It’s the Product Value 60 brainteaser from Illuminations:

Assign each letter a value equal to its position in the alphabet (A = 1, B = 2, C = 3, …). Then find the product value of a word by multiplying the values together. For example, CAT has a product value of 60, because C = 3, A = 1, T = 20, and 3 × 1 × 20 = 60.

How many other words can you find with a product value of 60?

CAT

As it turns out, there are 14 other words with a product value of 60. Don’t feel bad if you can’t find them all; while they’re all allowed in Scrabble™, the average person won’t recognize half of them.

You can see the full list and some definitions in this problem and solution PDF.

This problem resurfaced at the perfect time.

With 2016 just around the corner, no doubt many math teachers will present the following problem to students after winter break:

Find a mathematical expression for every whole number from 0 to 100, using only common mathematical symbols and the digits 2, 0, 1, and 6. (No other digits are allowed.)

And that’s not a bad problem. It gets even better if you require the digits to be used in order. For instance, you could make:

  • 2 = 20 + 16
  • 9 = 2 + 0 + 1 + 6
  • 36 = (2 + 0 + 1)! × 6

But that problem is a bit played out. I’ve seen it used in classrooms every year since… well, since I used it in my classroom in 1995.

So here are two versions of a problem — the first one being for younger folks — using the year and based on the Product Value 60 problem above:

How many words can you find with a product value of 16?

How many words can you find with a product value of 2016?

There are 5 words that have a product value of 16 and 12 words that have a product value of 2016 (spoiler: those links will take you to images of the answers). As above, you may not recognize all of the words on those lists, but some will definitely be familiar.

December 19, 2015 at 8:33 am Leave a comment

One, Tooth, Ree, …

Were a state assessment item writer to get his hands on the puzzle that I discuss below, I believe that this is what the problem might look like:

Fifteen congruent segments labeled A-O are arranged horizontally and vertically to form five squares, with five of the segments shared by two squares each. Which three segments can be removed to leave three congruent squares?

Toothpicks - Assessment Writer

  1. {A, B, C}
  2. {A, C, E}
  3. {B, K, N}
  4. {D, E, F}
  5. {G, I, N}
  6. {H, A, L}
  7. {I, B, M}
  8. {K, L, M}

You’ve likely seen this problem before, but in a simpler and more elegant form.

Remove three toothpicks to leave three squares.

Real Toothpicks

Eli shared this puzzle with us the other night at dinner, after hearing it during his math enrichment class that day. (It’s my sincere hope that every family has similar conversations at mealtime.) This was a great problem to discuss at the dinner table, since a collection of manipulatives was close at hand.

As you might expect, Eli and his brother started by removing toothpicks randomly and seeing what happened. But it turns out that random toothpick removal is not a great strategy. There are 15 toothpicks in the arrangement, so there are 15C3 = 495 ways to select three of them.

Of course, some of those selections are obviously wrong, such as selecting the three toothpicks in the middle row, or choosing the three highlighted in pink:

Toothpick - Remove 3 ToothpicksBut even ignoring the obviously wrong ones, that still leaves a lot of unobviously wrong combinations to consider. Yet I let my sons randomly select toothpicks for a few minutes without intervention. Why? Because of a primary tenet for effective problem solving:

Get dirty.

You can’t get any closer to a solution if you don’t try something. Even when you don’t know what to do, the worst thing to do is nothing. So, don’t be afraid to get your hands a little dirty. Go on, try something, and see what happens. You might get lucky; but if not, maybe it’ll shed some light.

After a few minutes, they conceded that random toothpick selection was futile. That’s when they modified their approach, leading to another important aspect of problem solving:

Look at the problem from a different perspective.

“Hmm,” said Eli. “That’s a lotta toothpicks.” He thought for a second. “But there are only five squares.” That’s when you could see it click, leading to the most popular problem-solving principle on the planet:

Solve a simpler problem.

There are 5 squares in the arrangement, and there are only 5C3 = 10 ways to select three of them. If you first choose three squares, then you can count the number of toothpicks that would need to be removed. For instance, you’d have to remove five toothpicks to leave these three squares:

Toothpicks - Possible Removal

That doesn’t quite work. But the problem is a whole lot easier if you focus on “leave three squares” rather than “remove three toothpicks.” There are only 10 possibilities to check.

From there, it was just a hop, skip, and jump to the solution.

So, maybe this problem, as well as the million or so other remove-some-toothpicks problems, isn’t very mathematical. It only contains two mathematical elements — basic counting and geometry. In the vernacular of Norman Webb’s Depth of Knowledge Levels, the kinds of questions that you could ask about this problem — “How many toothpicks are there?”, “How many are you removing?”, “What is a square?” — would be DOK 1.

But they’re fun, and they’re engaging to kids, and they develop problem-solving skills, such as those mentioned above as well as perseverance.

For more problems like this, check out Simply Science’s collection of 16 toothpick puzzles.

Speaking of the number 16, here’s my favorite toothpick problem.

Remove 4 toothpicks to leave 4 small triangles.

16 Toothpicks

And finally, because I know you came here looking for jokes, here is the only joke I know that involves toothpicks. It’s rather disgusting. Continue at your own risk.

One night, as a bartender is closing his bar, he hears a knock at the back door. It’s a math graduate student. “Can I have a toothpick?” he asks. The bartender gives him a toothpick and closes the door.

Five minutes later, another knock. “Can I have another toothpick?” asks the student. The bartender gives him another one and closes the door.

Five minutes later, a third knock. “Can I have a straw?” asks the student.

“Sure,” says the bartender, and hands him a straw. “But what’s going on out there?”

“Some lady threw up in the back,” says the grad student, “but all the good stuff is already gone.”

October 5, 2015 at 6:10 am 1 comment

You Say It’s Your Birthday

It’s my birthday, too.

Okay, not really, but it wouldn’t be surprising if it were. Today (September 17) is the fourth most common date on which to be born. Or at least it had been from 1973 to 1999 in the United States, according to an analysis by Amitabh Chandra, a professor of social policy at Harvard who compiled 28 years of birth date data.

Yesterday (September 16) was the overall most common birth date, and tomorrow (September 18) was the tenth most common.

In fact, all of the top ten most common birth dates occur in September, and even the lowest-ranking date of the month is still in the top third of the most common dates (September 5, #125). Apparently November and December are busy times for gettin’ busy.

It’s a busy birthday time elsewhere, too. Statistics New Zealand issued a warning about the excessive number of birthdays that will occur near the end of September.

How common is your birth date? Take a gander at the image below. Or, for an interactive experience, click around on Andy Kriebel’s version that uses Tableau.

Most Common Birthdays

This information (which is not new, btw — an initial analysis was completed by Chandra for a paper published in the Journal of Political Economy in 1999; a table with Chandra’s full data appeared in a column by David Leonhardt in 2006; and, various versions of the image above have been floating around the internet for several years) throws a wrench into the famous Birthday Problem, which asks, “At a party of n people, what is the probability that two of them share a birthday?”

The solution to the Birthday Problem assumes that birth dates are evenly distributed across the calendar. But clearly they’re not. How much does real data affect the solution, though? As it turns out, not much.

Using real birth date data from 1985-88 pulled from the CDC, Joe Rickert completed an analysis using Revelation R Enterprise 6. The results look like this:

Birthday ProblemOn the other hand, the theoretical probability of two people sharing a birth date in a group of n people is given by the formula:

\textrm{P} = 1 - \frac{365!}{(365-n)! \cdot 365^n}

And if you try to graph that formula, you get something like the following:

Birthday - Theoretical

Admittedly, I was lazy. The scatterplot above only includes n-values from 2‑10 and multiples of 5 from 15‑100. That’s because Excel can’t handle factorials larger than 170!, so I had to use a large number calculator online and enter the values into an Excel sheet by hand. Still, it gives a pretty good idea of the shape of the curve.

A typical benchmark for the Birthday Problem is n = 23, when the probability of a pair sharing the same birth date first exceeds 50%. Using real data, P(23) = 0.5087, and using the theoretical model, P(23) = 0.5073. Good enough for government work. By visual inspection, you can see that the other values match up rather well, too.

Finally, here are some math jokes about birthdays.

Statistics show that those who celebrate the most birthdays live longest.

I only drink twice a year: when it’s my birthday, and when it’s not.

Happy integer number of arbitrary units of time since the day of your birth!

You don’t need calculus to figure out your age.

They don’t make birthday cards for people who are 85 years old. So I almost bought you a card for an 80-year old and a 5-year old. But then I figured no one wants to do math on their birthday.

September 17, 2015 at 3:51 pm 1 comment

What’s Your Problem?

Problems in the MathCounts School Handbook are presented “shotgun style,” that is, a geometry problem precedes a logic puzzle and follows a probability question. (I worked for MathCounts for seven years and then served as a writer and chair of their Question Writing Committee, so I’m not unbiased.)

By comparison, textbooks often present 50 exercises on the same topic, each one only minimally different from the previous one. That tips the hand to students, methinks, and makes them realize, “Oh, I just need to do the same thing.” I prefer the MathCounts approach, where students have to dig into their bag of tricks to find a viable solution strategy.

With that in mind, here are a few problems I’ve encountered recently, each one not like the others.

Problem 1. The simple polygon is made from 73 squares, connected at their sides. What is the perimeter of the figure?

Thing Puzzle

A simple polygon made from 73 unit squares.

Problem 2. What is the expected number of times that a six-sided die must be rolled to get each number 1–6?

Problem 3. A wall is to be constructed from 2 x 1 bricks (that is, bricks that are twice as long in one direction as the other). A strong wall must have no fault lines; that is, it should have no horizontal or vertical lines that cut entirely through a configuration, dividing it into two pieces. What is the minimum size of a wall with no fault lines? The figure below shows a 3 × 4 wall that has both horizontal and vertical fault lines.

Fault Line - 3 x 4 Wall

A 3 × 4 wall with a fault line.

Please share great problems you’ve recently encountered in the Comments.


No answers, but here are some hints.

Problem 1. Look for a pattern.

Problem 2. Check out this simulation for the Cereal Box problem.

Problem 3. The smallest arrangement without a fault line is larger than 3 × 4 and smaller than 10 × 10.

December 1, 2014 at 6:54 am Leave a comment

All Beer and Skittles

The winter has been unkind to my waist, so I joined a month-long fitness challenge at the gym. I asked one of the trainers what I should do to win. “Drink 25 beers on Saturday night,” he said.

“Why?”

“Because the weigh-in is Sunday morning, and you want to be heavy as possible.”

“Yeah, I get that,” I replied. “But there are 24 bottles in a case. Okay if I stop there?”

How many is too many? Hard to say…

Teacher: If I had 5 bottles in one hand and 6 bottles in the other hand, what would I have?
Student: A drinking problem?

With St. Patrick’s Day just around the corner, here are a few more beer jokes.

If you pour root beer in a square cup, will it become beer?

And who doesn’t like a joke that makes fun of professors?

A literature professor, a computer scientist, and a mathematician head to a pub. They each order a pint of beer, and when the drinks are brought to the table, each pint has a fly in it.

The literature professor pushes her beer away in disgust.

The computer scientist removes the fly and proceeds to drink his beer.

The mathematician picks the fly out of his drink, too, but then holds it out over the beer and yells, “Spit it out, you little bastard!”

March 13, 2014 at 10:14 pm 2 comments

A 1-Derful Post for 1/1

My sons have refrigerator magnets with digits and binary operators, which they use to create expressions, equations, and dates. Recently, they created the following equation:

1 + 1 ÷ 1 – 1 × 1 = 1

They asked if it was correct. Oh, no, that’s not how things work in this house. “You tell me,” I said.

Eli said, “One plus one is two, divided by one is two, minus one is one, times one is one. It’s true.”

Anyone who teaches middle school has seen students make this type of order of operations error. The equation is true if operations are performed left-to-right but not if the conventional order of operations is applied.

On a calculator, I entered

1 + 2 ÷ 2

and asked, “What is the value of this expression?” Sure enough, they thought it would be three-halves, and they were surprised to see two displayed when the ENTER key was pressed.

Eli looked puzzled, and Alex looked cross. “Oh, right,” said Alex. “We have to do multiplication and division first.”

They then concluded that their equation was indeed true, and this time for the right reasons.

But it made me wonder:

What is the probability that the following equation will be true, if the four binary operators are randomly placed in the blanks with each operator used only once?

1 __ 1 __ 1 __ 1 __ 1 = 1

I’ll tell you that (a) I was surprised by the results and (b) I didn’t have to check every possible equation to arrive at the answer; in fact, I didn’t even have to check a quarter of them.

Surgeon: I have so many patients to see today! Who should I do surgery on first?
Nurse: Follow the order of operations.

How many calculus teachers does it take to screw in a light bulb?
0.99999999…

January 1, 2014 at 1:01 am Leave a comment

Making Progress, Arithmetically

Today is 11/12/13, a rather pleasant-sounding date because the numbers form an arithmetic sequence, albeit a trivial one. It’s not the only date in 2013 for which the month, date, and year form an arithmetic sequence. How many others are there?

Several nights ago, my sons asked if they could do bedtime math, but Eli asked if we could do problems other than those on the Bedtime Math website, because “they’re a little too easy.” So instead, I navigated to the MathCounts website and opened the 2013-14 MathCounts School Handbook. We scrolled to page 9 and attacked the problems in Warm-Up 1.

Things were going well until we reached Problem 8 in the set, which read:

The angles of a triangle form an arithmetic progression, and the smallest angle is 42°. What is the degree measure of the largest angle of the triangle?

Eli asked, “Daddy, what’s an arithmetic progression?” pronouncing arithmetic as “uh-rith-ma-tick” instead of “air-ith-met-ick.”

I could have just answered Eli’s question by stating the definition:

An arithmetic progression is a sequence of numbers for which there is a common difference between terms.

But such a definition isn’t very helpful, since I’m not sure that either Eli or Alex know what sequence, common difference, or term mean. It would have led to even more questions.

Plus, I’ve always believed that kids understand (and retain) more when they discover things on their own. Call it “discovery learning” or “inquiry-based instruction” or any of myriad other names from educational jargon, it just means that giving kids the answer is not the most effective way for them to learn.

So instead, I said, “Let me give you some examples.” And then I wrote:

1, 2, 3

3, 5, 7

Alex said, “Oh, I get it! An arithmetic progression is a nice pattern of numbers.”

So I said, “Well, let me give you some patterns that aren’t arithmetic progressions.” And then I wrote:

2, 4, 8

“That’s a nice pattern, isn’t it?” I asked. “But it’s not an arithmetic progression.”

“Oh,” said Alex. He thought for a second, then revised. “You have to add the same amount every time.”

And there you have it. Three examples, and my sons were able to define arithmetic progression. It’s not as sophisticated as “a common difference between terms,” but “add the same amount every time” is a sufficient definition for a six-year-old.

So they generated an arithmetic progression with 42 as the smallest term:

42, 45, 48

Eli said, “I don’t fink vat’s enough.” When asked to explain, he said he thought that the angles in a triangle add up to 180 degrees.

“Are you sure?” I asked. He wasn’t. Nor was Alex. So I asked if they could convince themselves that the sum of the angles is 180°.

Alex said, “Well, the angles in a square add up to 360°, and you could cut it in half.” So we did:

Square in Half

They then reasoned that each triangle would have a sum of 180°. “But maybe that only works for a square,” I said. “How do you know it’ll work for other shapes?”

Eli suggested that we could cut a rectangle in half, too:

RectangleHalfAnd again they concluded that each triangle would have a sum of 180°.

Understand, this is NOT a proof of the triangle sum formula. When they get to high school and need to demonstrate the rigor that the Common Core State Standards are demanding, well, then we’ll worry about formal proof. But for now, I’m okay with six-year-olds who can demonstrate that kind of reasoning.

They then took another guess, but this time they chose three numbers that added to 180:

42, 59, 79

Realizing that the difference between the first and second terms was 17 and the difference between the second and third terms was 20, they revised:

42, 60, 78

They concluded that the largest angle had a measure of 78°. And all was right with the world.

So why am I telling you all this?

Partially, it’s because I’m a proud father.

But more importantly, it’s because this vignette demonstrates that teaching is an art, and successful teaching doesn’t happen by accident. It’s not easy, as many people believe. What’s easy is the perpetuation of bad teaching, a la Charlie Brown’s teacher, or textbooks that simply present information with the belief that students will absorb it by osmosis. Good teaching, however, requires content knowledge and pedagogical knowledge, and it demands teachers who can handle unexpected classroom twists and turns and have the ability to adjust on the fly.

A student is convinced that a right triangle isn’t a right triangle because the right angle isn’t in the lower left corner? You better find an effective way to clarify that misconception. (Hint: Don’t use a traditional textbook where every picture of a right triangle shows the right angle in the lower left corner.)

Students think that 16/64 = 1/4 because you can “cancel the 6’s”? Uh-oh. Better find some counterexamples pronto, and help them understand why 16/64 can be reduced to 1/4.

Your students don’t know the definition of arithmetic progression? Then you better figure out a way to help them define it, and just writing your definition on the chalkboard isn’t gonna cut it.

Want to see what good teaching looks like? See Dan Meyer, or Christopher Danielson, or Fawn Nguyen. Or many, many others who don’t blog about it but inspire students every day.

Someday soon, I hope to add my project at Discovery Education to the list of examples of good teaching. Until then, I’ll just keep blathering about my sons.

November 12, 2013 at 10:54 am 5 comments

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About MJ4MF

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

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