Archive for February, 2015

Deliver Us Not Into Bad Math

What better way to celebrate National Pizza Day than sharing this sign, which hangs in our local Pizza Hut:

Pizza Hut Donation

Admittedly, I’ve never been very good with proportions, but even I know that

\frac{3}{14} \ne \frac{5}{30}.

Yet, that’s what’s implied by the statements for $3 and $5 in the sign. Further,

  • For $1, you can feed 4 children for 1 day. That’s a daily rate of 25.0¢ per child.
  • For $3, you can feed 2 children for 7 days. That’s a daily rate of 21.4¢ per child.
  • For $5, you can feed 1 child for 30 days. That’s a daily rate of 16.7¢ per child.

Will the real price per child per day please stand up?

And then I took a look at that last statement — that $10 can feed a classroom for a day — and it really blew my mind. Daily rates of 16.7 to 25.0¢ per child imply that classrooms have 40 to 60 students. I don’t know where these hungry students are, but maybe there should be a secondary campaign to reduce class size?

Though let’s be honest. What really seems to be needed here is an entirely new campaign:

Deliver Math Flyer

February 28, 2015 at 1:40 am 3 comments

Can I Get Your Digits?

Saw this on a t-shirt recently:

PIN Digits Pi
It made me think about this problem involving digits.

Consider the number


obtained by writing the numbers from 1 to 20 in order side-by-side.

What’s the greatest number that can be obtained by crossing out 20 digits?

If a fetching lady or handsome gent catches your fancy by solving that problem, you might want to ask her or him…

How can I know so many digits of π and so few digits of your phone number?

And if he or she still hasn’t taken leave of you, then you could really press your luck with the following:

  1. Ask your new friend to write down a number with four or more digits.


  1. Then, have your friend add the digits.

4 + 5 + 9 + 1 + 6 + 3 = 28

  1. Subtract the sum from the original number.

459,163 – 28 = 459,135

  1. Have your friend cross out one of the digits, and then read the remaining number aloud to you.


  1. Then, miraculously announce the missing digit.



The secret to the trick? Simple. Just add the digits of the number that your friend reads aloud, and then figure out what number must be added to get the sum to a multiple of 9. Above, the digits of the number 45,935 have a sum of 26, which is 1 less than a multiple of 9, so the removed digit is 1.


February 23, 2015 at 10:33 pm 6 comments

And the Oscar Goes To… a Mathematician?

Robert BridsonWhen you sit down to watch The Oscars on Sunday, February 22, you’ll be witnessing history when a mathematician takes home one of those 13.5″ tall gold statuettes.

Math professor Robert Bridson from the University of British Columbia will receive an Academy Award for Technical Achievement, in particular, for his “pioneering work on voxel data structures and its subsequent validation in fluid simulation tools,” according to the Academy of Motion Picture Arts and Sciences. Less technically, he figured out the mathematical software used to create breathtaking scenes for Gravity, Avatar, The Avengers, and The Hobbit.

Incidentally, Bridson is also the first person from Newfoundland to win an Academy Award. He said he hopes his win will inspire kids who like movies to consider a career in mathematics.

Bridson may be the first mathematician honored by the Academy, but his work is certainly not the first math to appear in a movie. The following gem, uploaded to YouTube by AlonzoMosleyFBI, shows 100 different movie clip in which each number 1‑100 is used in a quote. Awesome.

February 19, 2015 at 11:13 am 1 comment

Colored Pyramids and the Mind of a 7-Year-Old

This is what kept me up last night. Literally.

Form a row of ten squares, with each square randomly colored red, green, or yellow. Call this Row 1. Then place nine squares in Row 2 slightly offset above Row 1, and color the squares in Row 2 according to the following rules:

  • If two side-by-side squares have the same color, the square between them in the row above has the same color.
  • If two side-by-side squares have different colors, the square between them in the row above has the third color.

Continue in this manner with eight squares in Row 3, seven squares in Row 4, and so on, with just one square in Row 10.

Here’s an example of a pyramid constructed in this manner:

Color Squares Pyramid

The 1’s, 2’s and 3’s in the diagram appear because this image was created with Microsoft Excel. Formulas were used to determine the number in each square, and conditional formatting was used to color the squares.

So far, this is just a test of how well you can follow rules, and it isn’t much fun. But here’s where it gets interesting.

How can you predict the color of the lone square in Row 10 after seeing only the arrangement of squares in Row 1 (and without constructing the rows in between)?

That’s right…
Shit Got RealThis shit just got real.

I found this problem last night on the Purdue Math Department’s Problem of the Week website. I lay awake in bed longer than I should have, but no solution came to me either while laying there awake or while I was sleeping. When I woke up, I shared it with my sons. I suspected they’d have fun coloring squares; I never suspected what actually happened.

I explained the problem to Alex and Eli, and I showed them an example of how to generate Row 9 from an arbitrary Row 10. They then pulled out their box of crayons and constructed rows 8, 7, 6, …, 1. Alex then looked at his pyramid for about 8 seconds and said, “Oh, I get it. You can find the color of the top square by _________.” (Spoiler omitted.)

“Is that your conjecture?” I asked.

“What’s a conjecture?” he replied.

“It’s a guess,” I told him. “It’s what you think the rule is.”

“No,” he said. “It’s not a guess. That’s the rule.”

He was pretty cocky for a seven-year-old.

So we tested his rule for another randomly-generated Row 10. It worked. So we tested it again, and it worked again. We tested it for six different hand-drawn pyramids… and it worked for every one of them.

That’s when I generated this Excel file. We used it to test Alex’s conjecture on 100+ other pyramids. It worked every time.

I still have no idea how he divined the rule so quickly.

For me, though, the cool part came when I was able to extend the puzzle with the following:

For what values of k can you predict the color of the lone square in Row k when there are k blocks in Row 1?

Trivially, if I gave you just two blocks in Row 1, you could most certainly predict the color of the square in Row 2.

But if I gave you, say, five blocks in Row 1, could you predict the color of the lone square in Row 5?

The final part of the Puzzle of the Week description from the Purdue website says exactly what you’d expect it to say:

Prove your answer.

I have a proof showing that Alex’s conjecture holds. Incidentally, that proof can be extended to prove the general result for the extension just posed.

Alex, however, has not yet generated a proof of his conjecture.

Then again, he’s only seven.

February 15, 2015 at 8:34 am 5 comments

All You Need is LOVE

Valentine’s Day is almost here, but maybe you’ve been looking for love in all the wrong places. One possibility is to pop over to Wolfram Alpha and ask:

Will you be my Valentine?

Or, with a little mathematical creativity, you might be able to find some over at Desmos:

LOVE Graph

four equations were used to create this LOVE Graph

Or perhaps you’ve already found a special someone. If so, you might want to tell her how beautiful she is, using this (paraphrased) mathematical gem from Woody Allen:

Your figure describes a set of parabolas that could cause cardiac arrest in a yak.

(No, it’s not sexist of me to imply that readers would have girlfriends. It’s just that a compliment about a paramour’s curves doesn’t work so well when directed at a male.)

Perhaps your special someone makes your heart skip a beat.


Beating Heart graph, made from just one equation

If so, this graph can help you get your beat back:

Wherever you look for love on this Feast of St. Valentine, I hope you find it — or at least stumble on a couple of great problems to distract you.

February 11, 2015 at 2:14 am Leave a comment

Money-Saving Fermi Questions

I was pissed when my cousin wouldn’t give me two $5 bills for a $10 bill.

“Sorry, can’t,” he replied simply.

When asked why the hell not — I knew he had two $5 bills, because he had gotten one from the gas station attendant earlier, and the waitress just brought him another — he explained that all $5 bills are put into savings.

“When I receive a $5 bill, I don’t spend it. It stays in my wallet till I get home, and then it goes right into the piggy bank,” he said. “Every couple months, I take those bills to the bank. It’s an easy way to build up my savings account.”

“So, what, you save like $50 a year this way?”

“It’s a helluva lot more than you’d think,” he replied.

Five Dollar Bill

As stupid as this sounds, now everyone in my family is doing it. It is a low-impact way to build up your savings account. And it leads to a great Fermi question:

  • If all of your $5 bills go into savings, how much will you save in a year?

And for my sons, who don’t often pay for things with bills large enough to require $5 in change, we have the following:

  • If all of your nickels go into savings, how much will you save in a year?

Fermi questions are questions that require quantitative estimates to arrive at an answer. It often requires making assumptions, because exact data is unavailable. Here are a few others:

  • What percent of people who have ever lived are currently alive?
  • How many hot dogs are sold at Yankee Stadium during a baseball season?
  • How long would it take a snail to travel from Miami to Los Angeles?
  • What is the weight of a million dollars? (Assume 1,000,000 one-dollar bills.)

My favorite Fermi question is based on a Dunkin Donuts radio advertisement, in which they boasted:

We reject more than one million pounds of coffee beans a year.

Which has to make you wonder:

  • How picky are they, really?

February 4, 2015 at 9:59 pm Leave a comment

Solution to Super Bowl XLIX Problem

In the post When Super Bowl XLIX Starts to Bore You on January 29, you were asked to consider the following problem:

Imagine that the NFL has eliminated divisions, and there are just two conferences with 16 teams each. To simplify things, every team plays the other 15 teams in their conference exactly once each. At end of the regular season, what is probability that every team within a conference has a different record? (Assuming, of course, that each team is equally likely to win on any given Sunday, and assuming that ties are not possible.)

[UPDATE: Upon further review, it seems unlikely that you would have been bored during last night’s game. Unless you hate football and watched just for the commercials, which were terrible; or during the halftime show, which could be described as no better than fine; or during the post-game interviews, although if you’re like me, you were too busy retching over the Patriots’ victory to be solving a math problem just then.]

This is not a trivial problem, and a little exploration is good before you dive into a theoretical solution. Let’s start with a familiar heuristic, Solve a Simpler Problem.

If there were just two teams in the conference, one team would win the game they played. The winning team’s record would be 1‑0, and the losing team’s record would be 0‑1. That is, the records of the two teams would always be different, so P(different, 2 teams) = 1.

If there were three teams in the conference, it gets a little more complex, but not too bad. We can enumerate the outcomes of a season:

A > B
B > C
A > C
A > B
B > C
C > A
A > B
C > B
A > C
A > B
C > B
C > A
B > A
B > C
A > C
B > A
B > C
C > A
B > A
C > B
A > C
B > A
C > B
C > A
A: 2-0
B: 1-1
C: 0-1
A: 1-1
B: 1-1
C: 1-1
A: 2-0
B: 0-2
C: 1-1
A: 1-1
B: 0-2
C: 2-0
A: 1-1
B: 2-0
C: 0-2
A: 0-2
B: 2-0
C: 1-1
A: 1-1
B: 1-1
C: 1-1
A: 0-2
B: 1-1
C: 2-0


Of the eight possible outcomes, six result in the three teams having different records. That gives P(different, 3 teams) = 3/4.

Unfortunately, the number of possible outcomes jumps pretty quickly. With four teams, there are 64  possible outcomes, and I don’t have the time or energy to list them all here. You’ll either have to trust me, or you can write them out yourself. Of those 64, it turns out that 24 of them result in different records for all the teams. So, P(different, 4 teams) = 24/64 = 3/8.

By this point, you may have noticed a pattern. Or maybe not; it’s subtle.

So let’s start thinking theoretically.

Pick one of the teams. The probability that the first team you pick wins all its games is (1/2)15.

Then pick a second team. This team has one loss guaranteed, since the first team was undefeated. So, the probability that the second team wins its 14 remaining games is (1/2)14.

Then pick a third team. This team has two losses guaranteed, and the probability it will win its 13 remaining games is (1/2)13.

And so on.

The next-to-last team has a probability of (1/2)1 of winning its one remaining game.

And the last team has a probability of (1/2)0 of winning its zero remaining games. Which is a little weird, admittedly, but I include it here for completeness.

Now the kicker: There are 16! ways to choose the order of the teams.

So the probability of 16 teams each finishing with a different record is

16! \cdot \bigl(\frac{1}{2}\bigr)^{15 + 14 + 13 + 1 + \cdots + 0} = \frac{16!}{2^{210}} \approx 1.27 \times 10^{-50}

Which is to say, it’s not impossible… but nearly so.

February 2, 2015 at 7:07 am Leave a comment

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.

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.

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