What's Math Got to Do with It?: How Teachers and Parents Can Transform Mathematics Learning and Inspire Success (25 page)

One Solution

A good way to solve this problem is to look at how each part is growing separately. One way to see it is this: in the diagrams on the previous page, it looks like the white squares on the left are growing by one each time the pattern proceeds to the next step, as are the white squares on the right. The solid black square on the right and the solid black square on the bottom don’t seem to change. And the rectangle of gray squares is growing by one in each of its dimensions.
To come up with a formula, we need to number each picture. Let’s call the first picture “
n
= 1” and the next picture “
n
= 2.” So now let’s try to say how many of each square there are in terms of
n.
The white squares on the left and the white squares on the right always have one more than
n,
so these can each be represented by the expression (
n
+ 1). The black squares on the bottom and on the right are always just 1, no matter what
n
is, so these can be represented by the expression 1
.
Finally, the rectangle of gray squares has a width of
n
and a height that is two more than
n,
or (
n
+ 2)
.
So the number of squares in this rectangle can be represented by the width times the height, or the expression
n
(
n
+ 2). So the total number of squares on the
n
th picture is

(
n
+ 1) + (
n
+ 1) + 1 + 1 +
n
(
n
+ 2) =
n
+ 1 +
n
+ 1 + 1 + 1 +
n
2
+ 2
n
=
n
2
+ 4
n
+ 4

Interesting note: this expression factors as the perfect square, (n + 2)
2
, which means that every arrangement of squares can be arranged as a square. Try to see how they rearrange. This could lead to a different way of solving the problem.

The Amber Hill Question

Helen rides a bike for 1 hour at 30 km/hour and 2 hours at 15 km/hour. What is Helen’s average speed for the journey?

Solution

“Average speed” is one of those tricky expressions in word problems, because it can be interpreted to mean different things. The most natural interpretation is “If you were traveling at a
constant
speed, how fast
would you be going to cover the same distance in the same amount of time?” This interpretation allows you to work out the total distance traveled and the total time spent. Then the average speed is (total distance) / (total time). In this problem, Helen travels 30 km for the first hour and 15 × 2 = 30 km for the second and third hours. So the total distance is 30 km + 30 km = 60 km. The total time is 1 hour + 2 hours = 3 hours. So the average speed is (total distance) / (total time) = (60 km) / (3 hours) = 20 km/hour.

Chapter 7

The Staircase Problem

In this task students were asked to determine the total number of blocks in a staircase that grew incrementally from 1 block high, to 2 blocks high, to 3 blocks high, and so on, as a move toward predicting a 10-block–high staircase and a 100-block–high staircase, and, finally, algebraically expressing the total blocks in any staircase. Students were provided with a box full of linking cubes to build the staircases if they wished.

A four-block-tall staircase: total blocks = 4 + 3 + 2 + 1 = 10

Solution

There are many ways to “see” the growth in such a staircase. One of the most elegant is to think about two copies of the staircase fitted together, as shown here:

These fit together perfectly into a 4 × 5 rectangle. Since we put two staircases together, the total number of squares in our original staircase is (4 × 5)/2 = 10. In general, two of the
n
th staircases can be put together to make an
n
(
n
+ 1) rectangle. Again, since two staircases were put together, the total number of squares in one is
n
(
n
+ 1)/2. This is sometimes called the
n
th triangular number, because a staircase is triangular in shape. Using this formula we can work out how many squares are in each staircase. For example, the 10th staircase (
n
= 10) has 10(10 + 1)/2 = 55 squares. The 100th staircase has 100(100 + 1)/2 = 5050 squares. Rumor has it the famous mathematician Carl Gauss worked out in this way when his teacher asked the class to add up the numbers 1 through 100. Can you work out why this would give you the correct answer for that sum?

Alonzo’s Staircase Problem

5 blocks

5 + 9 = 14 blocks

5 + 9 + 13 = 27 blocks

Solution

Alonzo’s problem is like the staircase problem, except the staircase comes out in four directions from a center point. Again, we have numerous ways to count how many squares there are in this shape. One method is to use the solution to the staircase problem. Each of Alonzo’s staircases is formed by 4 copies of the staircases from the above problem, plus the squares in the center column, where the 4 copies are attached. So the number of squares is 4
n
[(
n
+ 1)/2] +
n
. The first term represents the 4 staircases, and the
n
is the column of squares in the middle. This last term is
n
because Alonzo’s shape is 1 high in the first case, 2 high in the second case, and in general
n
high in the
n
th case. This formula simplifies as follows:

4
n
[(
n
+ 1)/2] +
n
=
2
n
(
n
+ 1) +
n
= 2
n
2
+ 2
n
+
n
= 2
n
2
+ 3
n

Whenever you come up with a general algebraic formula, it’s good to make sure it works for the small cases that you understand. Can you try this formula to see if it works for the initial cases of Alonzo’s staircase pictured? Can you come up with this formula by grouping the cubes a different way?

Cowpens & Bullpens Problem

During an activity called “Cowpens & Bullpens” students had to determine how many lengths of fencing were required to contain an increasing number of cows, given certain fencing parameters.

Solution

There are many ways to see this problem’s pattern, as with others in the book. One way is to count the number of sides of fencing on each side of the cows, then add the corners. The first interesting thing to notice is that the number of sides of fencing above and below the cows is the same as the number of cows, and the number of sides of fencing to the left and right is always 1. So for the first picture above there are 4 + 4 + 1 + 1 + 4 = 14 sides of fencing, where the first two 4s come from the fencing above and below the cows, the 1s come from the fencing to the left and the right of the cows, and the last 4 comes from the 4 corners. For the next picture there are 5 + 5 + 1 + 1 + 4 for similar reasons. In general, there are
n
+
n
+ 1 + 1 + 4 sides of fencing, or 2
n
+ 6 sides.

Chapter 8
Problems from Sarah Flannery’s Book

The Two-Jars Puzzle

Given a 5-liter jar and a 3-liter jar and an unlimited supply of water, how do you measure out 4 liters exactly?

Solution

There are many ways to solve this problem—actually an infinite number! Here’s one way. Fill the 5-liter jar. Pour the water from the 5-liter jar into the 3-liter jar until the 3-liter jar is full. Now you have 2 liters of water remaining in the 5-liter jar. Dump out the water in the 3-liter jar. Then, put the 2 liters of water from the 5-liter jar into the 3-liter jar. Now fill up the 5-liter jar completely. Pour water from the 5-liter jar into the 3-liter jar until it is full. Since there were already 2 liters of water in the 3-liter jar, you have poured exactly 1 liter out of the 5-liter jar. So, there are exactly 4 liters remaining in the 5-liter jar.

If that was confusing, here’s a table showing how many liters of water are in each jar at each step, with explanations of what happened at each step: