What's Math Got to Do with It?: How Teachers and Parents Can Transform Mathematics Learning and Inspire Success (27 page)

8 = 4
+ 4 – 4

9 = 4 + 4 + 4/4

10 = 4 × 4 - 4 –

11 =
/4

12 = 4(4 – 4/4)

13 =
/4

14 = 4! – 4 – 4 –

15 = 4 × 4 – 4/4

16 = 4 × 4 + 4 – 4

17 = 4 × 4 + 4/4

18 = 4! –
–
–

19 = 4! – 4 – 4/4

20 = 4 × (4 + 4/4)

Notice for some solutions, I used the factorial notation, the exclamation point (!), after some numbers. This operation multiplies the number by every positive integer less than it. So 4! = 4 × 3 × 2 × 1 = 24. For the numbers where I used a factorial operation, do you think it’s possible to find solutions without it?

Solution

One of the things you may notice after playing this game for a while is that if you can get to 17, you are the winner. This is because no matter what your opponent adds, whether it be 1 or 2, on your next turn you will be able to get to 20. So getting to 17 is just as good as getting to
20. This idea can extend to even lower numbers. Here, 17 is a good number for you to get to because it is 3 away from 20, just one more than your opponent can add. Keeping 3 away from these “good numbers” is the trick. By similar reasoning, getting 14 makes you the winner, because no matter what your opponent adds, on your next turn you will be able to get to 17, and then you already know what you need to do to get to 20. Similar reasoning applies to 11, 8, 5, and so on, down by 3s. So now start at the beginning: Can you come up with a winning strategy if you go first? What about if you go second? If you like, make up some variations of this game that work differently, then try to find the strategy.

Painted Cubes

A 3 × 3 × 3 cube

is painted red on the outside. If it is broken up into 1 × 1 × 1–unit cubes, how many of these small cubes have three sides painted? Two sides painted? One side painted? No sides painted? What about if you start with a larger original cube?

Solution

The 3 × 3 cube has 1 cube right in the middle, with 0 sides painted. It has 6 cubes with 1 side painted, 1 in the center of each of the six faces. It has 12 cubes with 2 sides painted, 1 in the center of each edge of the large cube. And it has 8 cubes with 3 sides painted, 1 in each of the 8 corners of the large cube.

In general, for an
n
×
n
×
n
cube, let’s think about how to count all the cubes with 0 sides painted. Imagine removing the entire outer layer of small cubes. You’ll be left with a cube in the center, but each of its dimensions will be shrunk by 2, because a layer of cubes has been removed on both sides. So it’s an
n
− 2 by
n
− 2 by
n
− 2 cube. So it has (
n
− 2)
3
little cubes. For the cubes with 1 side painted, these are on the interior of each face. By similar reasoning as above, this is an
n
− 2 by
n
− 2 square, so there are (
n
− 2)
2
of these for each of the 6 faces, so 6(
n
− 2)
2
have 1 side painted. For the cubes with 2 sides painted, these are along each of the 12 edges, but there are
n
− 2 of these (can you see why?), making for a total of 12(
n
− 2) of these. Finally, no matter how large the cube is, there is only one cube with 3 sides painted per each of the 8 corners, so there are 8 cubes with three sides painted.

Solution

There are lots of ways to do this problem. One way is to break it into 11 cases, based on how many beans are in the first bowl. (Can you see why it is 11 cases and not 10 cases?) Then for each case, count how many ways the beans can be distributed among the other 2 bowls. I recommend doing this, so that you understand the problem more by immersing yourself in it. Once you have done that, here’s a quicker and more elegant way to see the final answer:

Imagine your beans as dots, all lined up, but with 2 extras:

• • • • • • • • • • • •

Why two extras? Well, imagine that it’s your job to replace two of the beans with an
x
, like this:

• •
x
• • •
x
• • • • •

or this:

• • • • • • •
x
•
x
• •

or even this:

• • • •
x
x
• • • • • •

These
x
’s are instructions on which bowls to put the beans into. They work as follows: put beans into the first bowl until you hit the first
x;
then put beans into the second bowl until you hit the second
x;
then put the rest of the beans into the third bowl. So in the first example above, there would be 2 beans in the first bowl, 3 beans in the second bowl, and 5 beans in the third bowl. For the second example, there would be 7 beans in the first bowl, 1 bean in the second bowl, and 2 beans in the third bowl. For the last example, there would be 4 beans in the first bowl, 0 beans in the second bowl (do you see why?), and 6 beans in the third bowl.