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.

National Math Storytelling Day!

Did you know that today, September 25 is Math Storytelling Day!  Yahoo recommends celebrating this quirky holiday with the following Abbot and Costello video clip:

In the clip, the donut baker (Costello) is convinced that 7 goes in to 28, 13 times.  In fact, in trying to convince Abbott he’s correct, he’s able to show this three different ways!

In each explanation, Costello misuses the idea of place-value to convince Abbott that 28 divided by 7 is 13.  But guess what?  From the mathematics he does, what he’s really showing Abbott is that 28 divided by 7 equals 1 + 3, which is exactly what we expect (and know) 28 divided by 7 equals!

Let me show you:

Costello’s first attempt at 7 into 28 is this, 2 can’t be divided by 7 (he says).  But, remember the 2 doesn’t represent 2 ones; it actually represents 2 tens (more commonly known as 20) so if he wanted to he actually could divide part of 20 .  .  . but we’ll get to that later.  Right now, its just important that you remember that the 2 actually represents 20.

Then, he divides 7 into 8, which is 1 with a remainder of 1 (he does this correctly).  So, he asks his sous chef to give him back the 2 (which is really 20) and writes “21.”  Now he proceeds to take 21 divided by 7, which is 3 and tells Abbott that 7 into 28 is 13, BUT 1 and 3 are both in the “ones” place meaning that 7 into 28 actually equals 1 + 3, or 4.

Costello really does this: Now, I have three questions for you:

1. Can you give similar explanations for the other two methods shown in the video?

2. If you were Abbott, how would you have convinced Costello he was incorrect?

3. Remember if 28 divided by 7 did equal 13, that means that we could take 28 things and divide them into 7 groups each containing 13 (in this case) donuts.  Suppose Costello really did take his 28 donuts and make 7 groups, each with 13 equal-sized donut pieces.  What fraction of a donut would each person get?