# Can you make 147?

Last week I was in a high school algebra classroom and the teacher was helping students get back into the swing of the school year and doing mathematics.  He was playing a game with the students in which he would display a number using a random number generator and the students would have to use all of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 as well as operations and grouping symbols to equal the number displayed on the screen.

For example, if the random number generator produced the number 17 a winning answer could be

17 = (4)²+1+((3+5+6+7+8+9)0) = 16 + 1 + 0 = 17

The first team to figure out a way to represent the number got 5 points.  If another team could represent the number in a different way they earned 2 points.  If a team checked another answer and found an error, they earned a point also.

The game was exciting!  It was loud!  People were yelling about the order of operations and exponents!  They were challenging each other about math!  It was music to my ears!

Then, the random number generator displayed the number 147.  It started off like any other round . . . whispering, writing, stopping, writing some more . . . and then something interesting happened–no one could figure out a way to combine the 10 digits to get the number 147.

I left that day thinking about the number 147 . . . was it impossible to generate the number using all 10 digits?  If so, how many other numbers could not be generated using this method also?  As I was driving back to my office I began thinking about the properties of 147.  First, I wondered is 147 prime?  But I quickly figured out that is was not prime (I used the test for divisibility by 3–1+4+7=12, which is divisible by 3, so 147 is divisible by 3).  The prime factors of 147 were 7 and 3, in fact 147 = 7²(3).

Now I had a strategy!  Was there a way I could combine 10 digits to equal 7, 7, and 3?  If so I could represent 147 in the way that was so tricky!  Guess what?  I could!  And now its your turn!  Show me how you represented 147!  Can you do this in a way that doesn’t build off of the prime factorization of the number?  Good Luck!

Want to recreate this activity yourself?  Here’s a random number generator.  Want to learn more about divisibility tests?  Look here. And you can find a related post here. Or here.

# In like a Lion; Out like a Lamb?

This morning as I was walking to work through whipping wind and freezing cold temps, I thought to myself “March, you’re supposed to be going out like a lamb.”  Goodness knows it sure came in like a lion!  Then, I started thinking about Groundhog’s Day and how unreliable that good old Phil actually is!  That made me wonder . . . is there really any truth to this whole lion/lamb thing?

I spent the better part of my morning trying to track down an answer!

This is what I did:

I’m only concerned with how March comes in and out, and not what happens in the middle of the month, so I thought I’d look at the first 7 days of March and the last 7 days of March.  Then, I did a quick search and found that the average March temperature in Iowa for the past 150 years is 34.5 degrees Fahrenheit.

Given this information I decided I would define “Lion” to be a 7-day span in which 4 or more of the days had an average temperature that was less than the average monthly temperature.  Then, a “Lamb” was a 7-day span in which 4 or more of the days had an average temperature that was greater than or equal to the average monthly temperature.

I obtained daily average temperature data for Des Moines from The University of Dayton Average Daily Temperature Archive dating back to 1995.  Because the month of March isn’t over yet for 2014, I didn’t use the temperature data for any of the days in March 2014.  This is what I came up with:

In the past 19 years March has followed the “In like a Lion; Out like a Lamb,” pattern in 13 different years (or 68% of the time).  It has followed an “In like a Lamb; Out like a Lamb” pattern 5 different years (or 26% of the time).  And once in the last 19 years March has come in like a Lion and gone out like a Lion . . . according to the average daily temperature in Des Moines.

This got me thinking about a few things:

1. This saying seems to be a little more accurate then the Groundhog shadow thing.

2. There weren’t any “In like a Lamb; Out like a Lion” years . . . I wonder how many times that has happened in the past 150 years (if at all)?

3. What do you think of my definition for Lion-like weather and Lamb-like weather?  Would you define it another way?  If so, how?

# Math Dice Game

My mom bought me this little game called “Math Dice” for Christmas this year.  Have you heard of it?  I hadn’t, and truth be told, I’m not sure my mom realized it was a game she was buying it for me.

When I opened the package she said “I thought you could do something creative with those dice and your math blog.”  In the days after Christmas, I scooped the unopened box of dice into our “junk drawer” (sorry mom) and rediscovered them this weekend while cleaning.  On the back the of the box were the directions to the “Math Dice Game.”

This morning I had a little extra time, so I thought I’d give the game try!

Step One: Roll the 12-sided target dice

Step Two: Roll the three 6-sided scoring dice.  Combine the three scoring dice in anyway to match or come closest to the Target Number.

Ummm, really?  1, 1, 2?  The closest I could get to 80 was 24.  This is what I did:

(1+1+2)! = 4*3*2*1.  Can you get closer?

My second roll of the Target Dice was 36

My Scoring Dice roll was 6, 4, 2

Super easy: (6*(4+2))=36.  Did you get 36 another way?

My last roll was 20

And my scoring dice were 6, 3, 1

I couldn’t get 20.  I could get 18 and 21, but 20 right on the money was a little tricky.  Can you do it?

Much, much more to come about this fun Math Game, with my new Math Dice . . . I’m working on a table of possible Target Number combinations as we speak!

# 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!

# And the Winner is . . .

So, did you see that last week some lucky person in South Carolina won the \$400 million Powerball Jackpot? (Technically, it was \$399.8 million, but \$400 million is close enough!)

It seems like every time there’s a big jackpot won in the lottery you read statements like “you’re more likely to be struck by lightening,” or “you’re more likely to marry a prince,” or in the case of the CNN story about this particular Powerball winner “you’re more likely to get struck by lightening and bitten by a shark.” (Talk about a bad day!)  They also go an to say that the chances of winning a Powerball Jackpot are 1 in 175223510.  Don’t you wonder how people come up with all of these statistics?

Let’s take a look at how Powerball is played . . .

According to the Powerball website, lottery numbers are drawn from two drums.  The first drum contains 59 white balls and the second drum contains 35 red balls (These red balls are all potential Powerballs).  The jackpot is won by matching all five white balls in any order and the red Powerball.

In order to calculate the odds of winning, we need to figure out the odds of matching all five white balls, in any order, and the red ball.

I like to think about situations like the one described above by picturing an empty (in this case) lottery ticket.  Like this:

The first ball that pops up could be any of the 59 balls, the second ball could be any of the 58 balls, the remaining blank spots on the ticket will be filled by drawing from the final 57, 56, and 55 balls respectively.

That looks like this:

Now, the order doesn’t matter in the way I arrange the balls, remember?  That means if the winning white balls are 1, 2, 3, 4, 5 and my ticket is 2, 3, 4, 1, 5; I’m on my way to winning the Powerball!  So now we need to figure out how many different ways the 5 white ball numbers can be arranged.

The first white number could be any of the 5 numbers drawn from the drum, so I have 5 choices for the number in the first spot.  I only have 4 remaining numbers for the second spot, 3 for the 3rd spot, etc . . .

Because I can rearrange the numbers and still have a winning ticket, the possible number combinations I need to win has just been decreased!  Now, we can calculate the number of ways to get a winning combination from the white balls in the Powerball drawing:

The total number of combinations is :

And now for the red Powerball!  The process we used above is going to be the same for red Powerball, except instead of having to match 5 numbers you only have to match 1 and because there’s only one number to be matched it doesn’t really make sense to talk about whether or not the order matters . . . there’s only one number.

Since you know a method to use and you know the answer, I’m pretty confident you are going to be able to figure how CNN could report that a person has a 1 in 175223510 chance in winning.

Good Luck!

P.S. I used a few different techniques from Discrete Mathematics or Counting Theory in this post that I didn’t explicitly name.  First, as in the case of the white lotto balls I was choosing 5 balls from a collection of 59 balls.  The order in which I arranged these 5 balls was not important.  This situation describes a combination or a binomial coefficient.  There is a formula associated with these types of situations.  We used this formula, although I did lots of canceling to make the numbers used look less overwhelming.  You could rewrite the situation that I described above as a combination using the formula provided from the combination link and see if you can get it to look like the one I used.

Also, we calculated the factorial of 5, denoted 5!.  If you attempt the challenge I’ve given you above, you’ll want to make sure you know what a factorial is.