# mwittels/khan-exercises forked from Khan/khan-exercises

Redo hints to be better and add some color

ss3 committed Aug 9, 2012
1 parent a201966 commit 4d249b5bc65850997cae1dff36dccfeec4de63b5
Showing with 20 additions and 21 deletions.
1. +20 −21 exercises/pile_patterns.html
 @@ -1,5 +1,5 @@ - + Pile patterns @@ -177,44 +177,43 @@
In step 0, we have START blocks. At each step, we add XS more blocks.

- So, at step number S, we have
XS \times S + START blocks. + So, at step number S, we have
START + XS \times S blocks.
-

The number of blocks in the diagram make an arithmetic progression. This means that at each step, the same number of blocks are added. We want to find a way to figure out how many blocks there would be at any step.

-

First, we need to find out how many blocks there are at the beginning. To do this, look how many blocks there are in step 0 by dragging the slider all the way to the left.

+

The number of blocks in the diagram make an arithmetic progression. This means that at each step, the same number of blocks are added. We want to find a way to know how many blocks there will be at any step.

+

First, we need to find out how many blocks we start out with. Count how many blocks there are in step 0 by dragging the slider all the way to the left.

-

We see that there are START blocks in step 0. Fill this in the answer box.

+

We see that there plural("is", START) pluralTex(START, "block") in step 0. Fill this in the answer box.

graph.slider.moveTo(1, 0);

Next, let's figure out how many blocks are added in each step. The easiest way to do that is to see how many blocks are added from step 0 to step 1.

-

There are START + XS blocks in step 1. To find how many blocks are added in each steps we just subtract this number from the number of blocks in step 0. So there are START + XS - START = XS blocks added in each step. Fill this in the answer box as well.

+

There are START + XS blocks in step 1. To find how many blocks are added in each steps we just subtract this number from the number of blocks in step 0. So there are START + XS - START = XS plural("block", XS) added in each step. Fill this in the answer box.

graph.slider.moveTo(3, 0);

Last, we have to figure out how many blocks there would be at any step. We can answer this by thinking "If we are at some step S, how many blocks would there be?"

-

Because this is an arithmetic progression, we can represent the number of blocks using an expression that looks like: a \times S + b where a and b are numbers that we have to fill in.

+

Well, let's start looking at how many blocks there are, and we'll see if there's a pattern.

+

Step 0: pluralTex(START, "block")

-

If we are at step 0, we can figure out what b should be. We know that there are START blocks in step 0. So let's plug into the equation:

-

a \times 0 + b = START

+

We know step 1 has START + XS blocks, but we can also write it as the number of blocks in step 0, plus the number of blocks we add from step 0 to step 1:

+

Step 1: (Number of blocks in step 0) + XS = START + XS blocks

+

We can do the same thing with steps 2 and 3, because we know we add the same amount of blocks in each step:

+

Step 2: (Number of blocks in step 1) + XS = START + XS + XS blocks

+

Step 3: (Number of blocks in step 2) + XS = START + XS + XS + XS blocks

-

Since 0 times anything is 0, we can turn this into:

-

0 + b = b = START

-
-

Now that we know that b = START, we can figure out what a is by using what we know about step 1.

-
-

Plugging in again, we get:

-

a \times 1 + START = START + XS

-
-

By solving for a, we find that a = XS. Do you see any similarities between what we got for a and b and the numbers we found above?

-
-

That's right, b is just the number of blocks at the beginning, and a is just the number of blocks we add in each step.

-

We now have the answer. In step S the number of blocks is XS \times S + START

+

Here, we can see a pattern! The number of blocks in each step can be represented as the sum of pluralTex(START, "block") and some number of groups of pluralTex(XS, "block"). How many groups of pluralTex(XS, "block") are there in each step?

+
+ $(".const").addClass("hint_blue"); +$(".lin").addClass("hint_orange"); +
+

That's right, the same number as the step they are on!

+

So, we can combine the groups of blocks together to look something like START + XS \times S, where S represents which step we are on.