Skip to content

AFL scores

October 31, 2012

The AFL season is nearly upon us again.

In this spirit, a small problem, but still a nice one, which some of my students may already know.

It is solvable either by trial and error, or using algebra…

What AFL scores have the number of points equal to the product of the goals and behinds?

For example, 4g 8b = 32 points, and 4 x 8 = 32.
On the other hand, 10g 15b = 75 points, but 10 x 15 ≠ 75.

See AFL scoring if you’re unfamiliar with the game.

I’ll give you some thinking space (scroll down for a hint)…

A hint: There are only four such scores, all with less than 50 points as the total.

Another hint: Pick a (small) number of goals and try to get a number of behinds that matches – excuse the pun :-).

Some more thinking space (scroll down some more for more hints)…

You could try, say

3g 3b = 21 points, which is 12 more than 3×3 = 9
3g 4b = 22 points, which is 10 more than 3×4 = 12
3g 5b = 23 points, which is 8 more than 3×5 = 15
3g 6b = 24 points, which is 6 more than 3×6 = 18

When will the difference shrink to zero?

Then what about other numbers of goals?

Here are the answers…

2g 12b = 24 points (and 2×12 = 24)
3g 9b = 27 points (and 3×9 = 27)
4g 8b = 32 points (and 4×8 = 32)
7g 7b = 49 points (and 7×7 = 49)

You may have experimented your way to the above solutions
If so, you may – quite rightly – have a hunch that there are no more.
As I sometimes say to students, don’t trust your hunches, but do listen to them.
They’re probably at least worth checking up on.
How do we prove that there are no more?

Let’s use the following names:
G = number of goals scored
B = number of behinds scored
P = total number of points scored

Then
P = 6 x G + B
P = B x G

So, putting the P’s together
B x G = 6 x G + B

Move the B on the right hand side to the other side of the equation
B x G – B = 6 x G

Take out B as a common factor on the left hand side
B x (G – 1) = 6 x G

Divide both sides by (G – 1)
B = 6 x G ÷ (G – 1)

Now, B must be a positive whole number, so (G – 1) must be a factor of 6 x G
(that is, dividing 6 x G by (G – 1) must leave no remainder)

(G – 1) is not a factor of G unless (G – 1) = 1 – try it for different values of G.
So, either (G – 1) = 1, or (G – 1) must be a factor of 6.
So (G – 1) can be 1 (for both reasons), 2, 3 or 6.

So the only solutions are, using B = 6 x G ÷ (G – 1):

(G – 1) G B P
1 2 12 24
2 3 9 27
3 4 8 32
6 7 7 49

Notice that each team can have no more than one of these scores during the course of any particular game.

Advertisements
Leave a Comment

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: