Monthly Archives: October 2013

Reblogged from The Atlantic: The Myth of I’m Bad at Math

Today I’m reblogging a post I saw in The Atlantic this morning.  I’m reblogging it because I think the authors, Miles Kimball and Noah Smith, do a great job of articulating a sentiment that I think many, many mathematics teachers share.

(And because they talk about Terence Tao, and any true Math Warriors fan knows his importance in the world of mathematics!)





This is the second post in a trilogy!  Read the first post here.  And when you’re done with this post, read the last post here.

Have you been counting chocolate?


Did you eat the chocolate you counted?

Even better!

If we pick up where we left off last week, you’ll recall the M and M debacle (the debacle being the fact that my children fought, not the fact that I fed them chocolate before 8am)!

You remember the fight, don’t you?  You know, the one where my daughter said it wasn’t fair that her brother got all the colors of the M&M’s in his bag and she didn’t?

Then, I asked you to try to figure out how likely that situation was to occur?  And then . . . I asked you to buy lots and lots of M&M’s?

Well, did you?  I did!

See . . .


The question that led to all of this M&M consumption was “How likely is it that a bag of fun sized M&M’s will contain exactly two colors of M&M’s?”

I asked you to start by looking at a sample of M&Ms.

My sample contained 17 bags of fun sized M&M’s and 299 candies.    When collecting the data to answer this question, I did two things.

1) Counted the total number of M&M’s in the bag.

2) Counted the number of each color of M&M in the bag.

My piles looked like this:

photo copy photo

And the data I collected looked like this:

Screen Shot 2013-10-25 at 1.33.28 PM

If you look closely, you’ll see that many times during the data collection phase I opened a bag of M&Ms that was missing one color.  In fact, this particular event occurred approximately 35% of the time.

Screen Shot 2013-10-25 at 1.36.03 PM

And when a color was missing, 33.3% of the time it was the blue or brown M&Ms that were missing and 16.67% of the time it was the orange or yellow M&Ms that were missing.

In my study, two or more colors were missing 0% of the time.

Considering this data, it seems very, very unlikely that my daughter would have opened a bag of M&M’s containing only two colors of M&M’s.  In fact, in my data collection this occurred 0% of the time.

So, here’s my next question; since the instance my daughter described happened 0% of the time during my study does it mean it will never happen?  How do you know?

Could I use this data to make an educated guess that my daughter my have just been picking a fight with my son, and that her bag of M&M’s did, in fact, contain more than just two colors of M&M’s?

To be continued . . .

M and M’s

This actually ended up becoming a 3 part series . . . all devoted to M&M’s!.  This is the first post in the series, but you can read the other two here and here.

This morning a box of fun-sized M&M packages got delivered to our house (thanks to my mother-in-law aka Grandma).  Although this seems like a thoughtful gesture, its a major problem.  Why?  You might ask?  Easy.  I have absolutely no will power when it comes to M&M’s, so much so, that a package of M&M’s may have been opened at breakfast this morning.  And a package may have also been opened for my 7 year-old.  And my 5 year-old.  And my 2 year-old.  (Don’t be impressed with my husband, the only reason he didn’t open a package is because be was already at work!)

Chocolate in the morning.  Everyone should be in a good mood, right?  Even better is chocolate covered in brightly-colored candy coating.  Unless, of course, that candy coating happens to lay the groundwork for a fight between two pint-sized humans.

“Mom! Jack’s package of M&M’s has all the colors and mine only has two!  That’s not fair!  Tell him that’s not fair!’

Hmmm, I’m not sure that “not fair” is the phrase that should be used to describe this particular incident.  How about “Mom!  That’s not likely!”

I know, I know “not likely” is not nearly as moving as “not fair,” but this is a blog about mathematics after all.  Fairness is not something that I can speak to in this context, but likely, now that’s something we can talk about!

Here’s what I want you to do.  And you have to promise to not cheat.  (Promise).  Buy a few bags of M&M’s this weekend.  As many as you see fit and go ahead and try to determine how “likely” it is that someone will get all the colors of the M&M’s in one fun sized bag.  Also, see if you can’t figure out if M&M Mars makes equal numbers of red, orange, yellow, green, blue, and brown M&Ms.

Remember, you already promised you wouldn’t cheat.  So I’m trusting you not to just Google this as soon as you’re done here.  Besides, you’d get to eat the project when you’re done.  If you just Google it, there won’t be anything good to eat.

I’m going to answer this question too.  And I’ll show you how I went about answering it . . . next week.  In the meantime, send me your methods.  Let’s see what we can come up with!

Lowering the Average Age at Panera . . .

This morning I was tasked with bringing a tote of Panera Hazelnut Coffee to work; which was fine with me because it was the perfect excuse to have a cinnamon crunch bagel for breakfast! (I actually tried their pumpkin pie bagel this morning, which was new to me . . . I don’t know if its new the store! Oh. and I’ll be getting it again on Sunday . . .it was that good!)

Anyway, this morning’s trip to Panera was not a solo mission.  My 3 children (ages 7, 5, and almost 2), as well as my husband (age 33) and myself (age 32) burst through the doors to Panera at 7am (we did literally “burst” if you’re a parent you know, and if not just trust me).

Here’s the thing about taking 3 small kids out to eat; we pretty much draw attention to ourselves whether we’re grabbing a quick breakfast (like this morning) or if we’re living on the edge and attempting a sit down dinner at a niceish restaurant.  Don’t get me wrong, I have good kids (most of the time), but its just that their excitement about little things tends to gain most people’s attention!  Here’s the breakdown for the way we are usually received:

Grandparent-aged people: smile, sometimes talk to the kids, usually say something along the lines of “what a nice family” to my husband and I.

Middle-aged people: debatable, sometimes smiley, sometimes they try to ignore us, sometimes in rare cases they move away from us.

Parent-aged people: they don’t even know we’ve entered the restaurant; they’ve got cats of their own to herd!

Anyway, this morning as soon as we walked into Panera it was all eyes on us.  My children were stars!  Why? you might ask.  Easy.  Every other customer in the entire restaurant fell into the “grandparent-aged” category and were making over our children!

When we finally sat down to eat our breakfast I looked around and thought “I wonder what the average age of people in the Panera at 7am is?” and then, I tweeted this:

Screen Shot 2013-10-11 at 11.58.21 AM

But, if I’m being honest with you.  That tweet has been bugging me all morning, because I wonder if its really true.  Did we lower the average age of the customers at Panera by 20 years?

First, I’m going to go ahead and assume that everyone else in the restaurant at that time was retirement age.  I’m also going to assume that there were about 30 people (not including us) in the restaurant.  According to a couple of different sources, the average age of retirement is 62. I would venture a bet that most of the people at Panera this morning were older than 62, but we’ll stick with 62.  OK, so here’s my question:

Did the 5 of us (ages 1, 5, 7, 32, 33) really lower the average age at Panera by 20 years?

Unbelievable Math Problem

During my first year of teaching one of my Algebra I students printed out this chain email and brought it to me to solve:


I don’t remember the exact conversation, but I do remember that it was something along the lines of

“My mom and sister and I spent the weekend trying to do this and we bet you can’t figure it out!”

Just in case you can’t read the problem in the photo, I’ll re-type it for you (but I have to admit I added the missing capitalization and punctuation . . . I just couldn’t leave alone!):

Wow, this is spooky.


Here is a math trick so unbelievable that it will stump you.  (Or at least it stumped me and I have a degree in math!)  Personally I would like to know who came up with this and where they had the time to figure this out.  I still don’t understand it!

1. Grab a calculator (you won’t be able to do this one in your head).

2. Key in the first three digits of your phone number (NOT the area code).

3. Multiply by 80.

4. Add 1.

5. Multiply by 250.

6. Add the last 4 digits of your phone number.

7. Add the last 4 digits of your phone number again.

8. Subtract 250.

9. Divide the number by 2.

Do you recognize the number?

*Disclaimer about this email . . . I was very motivated to explain to the student how this worked, because I’m not a fan of the whole “math is incredibly complicated and hard to figure out,” movement that seems to sweep across quite a few mathematics classrooms.  (We’ve talked about that in my “Calculating Tips is Calculus” post).

Anyway, let’s get to the bottom of this so-called “math trick.”  If you haven’t already go ahead and follow the steps in the email.  If you’re doing this on a calculator, as the email suggests be sure to press “enter” or “=” at the end of each step.

So, what happened?  Do you recognize the number?  It’s your phone number, right?

Hmmm, I wonder if this works with everyone’s phone number?  Try your mom’s phone number, or your gramma’s phone number, or your best friend’s phone number . . . does the trick work with their numbers too?

Well that’s tricky, isn’t it? (or is it?!?)

Let’s start at the beginning of the math trick email.  You start by entering the first 3 digits of your phone number.  Now, we know this trick already works for your phone number, and the other phone numbers you tested, but let’s see if we can figure out why/if it works for all phone numbers.  Instead of a specific number, I’m going to say the first 3 digits of my phone number are:


This is important, do you know why I’m doing this (this being symbols instead of numbers)?

No?  Let me tell you; If I pick the numbers 123 as the first three digits of my phone number, I’m not explaining why this math trick works for all phone numbers.  I’m only showing that this math trick works for phone numbers that have 123 as their first 3 digits.  If I wanted to assign the first 3 digits actual numbers, I’d have to test this math trick for all possible combinations of the first three numbers!  That a total of 729 different first three numbers . . . that’s a lot of “math trick” tests (and it doesn’t even include the different combinations of first 3 digits, plus last 4 digits . . . that’d be a total of 4,782,969 different combinations to check this math trick for).  So while it seems funny to use symbols, since the symbols can represent any first 3 digits it ends up saving me a lot of time in the long run.

So, step 1: Grab a calculator (I won’t be able to use a calculator if I’m going to replace the digits with symbols, so I’ll grab a piece of paper instead).

Step 2: Instead of keying in the first 3 digits of my phone number, I’m going to use the symbols.

Step 3: Multiply by 80: 80(#$%)

Step 4: Add 1: 80(#$%) + 1

Step 5: Multiply by 250: 250[80(#$%)+1)]–OK at this point the expression is getting a little ugly, so let’s go ahead and use the distributive property to simplify this mess!  When doing that, I get:


Step 6: Add the last 4 digits of your phone number: I’m going to use ^&*@ as the last 4 digits.  This means now I have:

20,000(#$%)+250+^&*@–I know, I know its getting a little weird just stick with me!

Step 7: Add the last 4 digits of your phone number again: OK at this point I’m adding another round of ^&*@, giving me 2^, 2&, 2*, and 2@ so, I’m writing this expression as:


Step 8: Subtract 250: Thank goodness, that 250 was getting to be annoying!  Now you have

20,000(#$%) + 2(^&*@).

Step 9: Divide by 2:  Well, what do you know?!?  The coefficient of #$% and ^&*@ are both divisible by 2!  That means I’m left with

10,000(#$%) + ^&*@

That’s a phone number!  Look, you take the first three digits of a phone number and multiply by 10,000.  Remember, multiplying anything by a power of 10 just moves the decimal point, so 10,000(#$%) = #$%0000 and what do you know; when I take #$%0000+^&*@, I get #$%^&*@–which is my phone number!

But, I have a challenge for you: Change the math trick so that it works if the person includes their area code.  When you come up with one, let me know!  I’d love to see it!

They’re baaaaaaack!

Have you seen the web series Math Warriors?


No?  If you haven’t, there’s good news!  You have enough time to catch up on the first 3 seasons, before the 4th season starts any day now!

Yes?  Then, I have bad news.  You have to wait a few more days before the 4th season begins!  (Although I guess you could rewatch a few of your favorite episodes).

The Math Warriors are a team of students from Yale competing in math competitions.  In the first two seasons they end up having to face their arch rival Harvard, although last season they crossed the pond to compete against the team from Wales.

Anyway, most of the series follows the team as they get ready to compete (and Felcia as she vies for the Miss America title . . .just watch and you’ll understand).  The team leader, Felcia, is pretty fanatic about two things: Prime Numbers and the Right Hand Rule.

We’ve talked about Prime Numbers on this blog before, so I’m going to assume that you know what they are.  (Although if you don’t, Felcia’s mom does a great job of explaining in Season 1, Episode 1!)  But I thought you might be a little less familiar with the Right Hand Rule.

In anticipation of Math Warriors, Season 4 I thought we’d consider the Right Hand Rule.

The Right Hand Rule is used as a way to determine the orientation of the cross product of two 3D vectors.  Remember, a vector is a ray that has both a magnitude (length) and a direction (north, south, east, west, or NW, NE, SW, etc., etc).  The result of the cross product of two vectors in 3D is always a vector that is perpendicular to the two starting vectors.  While the Right Hand Rule won’t help you out with the arithmetic needed to calculate the location of the cross product, it is a handy little tool used to help you visualize the direction of the cross product.

Want to see the Right Hand Rule in action? This physics professor has a really nice explanation of the Right Hand Rule (with pictures!)

Now, who’s ready for Math Warriors Season 4!  Right Hand Rule!