Image courtesy Google Images, original painting by Leslie Stones.
Here is a typical problem from a generic math competition: Farmer Brown has pigs and hens in a pen. If he sees heads and legs, then how many pigs are in then pen?
We've all been taught how to do this in school. Let be the number of hens, and let be the number of pigs. We see that we have the following system:
Then we'd maybe divide the second equation in half to get , subtract the first from this new equation to get , and we'd be done. Same ol' same ol'.
However, if you have ever been to a math competition before, you have probably seen some people do these problems. Just after giving out the number of hens and pigs, they can immediately give out an answer, before you can even write out the system. How do they do it? Do they memorize? Is there some sort of formula?
If you have ever thought these thoughts before, then you're in luck; today, I will be talking about shortcuts in solving these linear systems.
Step 1: Assume there are only pigs, or only hens. Well, let's assume there are only hens. Since there are heads, there will be legs. However, we've fallen short! This is where the trick comes.
Step 2: Calculate how many more legs would be there if instead of a hen, there was a pig. This is simple; since before, there will be legs, and after, there will be legs, there is a -leg increase.
Step 3: Calculate how much of this increase we need. Since we want legs, and we only have , we need more legs.
Step 4: Solve. Since replacing one hen with one pig will give us more legs, we need to replace hens with pigs to increase the amount of legs by . We are done now; there are pigs, and hens.
You may think that this is a lot more work than just setting up the system of equations and solving like that, but try it yourself. You'll be surprised by how fast you can do it with practice.
Before we dive into the details, here is a practice problem for you: A dark attic houses spiders and flies. If there are heads and legs in the attic, how many spiders are inside?
Let's look at the system of equations again: What did we do? Well, at first, we assumed that all the animals were hens. We can represent this by pretending that the pigs had two legs: We can see that should replace the . This corresponds to our first step.
Now obviously, , so we have to continue solving. What did we in step ? Well, we calculated how many more legs we needed. So why not do the calculation now:
Aha! We can see that the corresponds to how many more legs per switch, and the corresponds to how many more legs we need. Now that we are done with step and , we get get on with step , which should be obvious now.
We simply divide by from both sides to get , and we are done rigor-fying our procedure.
This procedure, when shown mathematically, seems no simpler than the classic elimination procedure, but the real power comes when this trick is put into play. Before you know it, you'll have mastered this trick and people will be asking you why you're so good at math.
*1. * On the planet Boorg, there are three-armed inhabitants and five-armed inhabitants (one head per inhabitant as usual). If in a house there are heads and arms, then how many five-armed inhabitants are there? How about three-armed inhabitants?
*2. * I have some cent stamps and some cent stamps. If I have a total of stamps and cents, then how many cent stamps do I have? How many cent stamps?
*3. * There are some more spiders and flies in an attic. If there are heads and legs, how many flies are there? Did the trick work? Why or why not?
*4. * Why is it easier to calculate how much of the thing with the larger amount of whatever they have, let it be cents, arms, or legs, than the the thing with the smaller amount? In other words, why is it easier to calculate how many cent stamps there are than cent stamps?
*5. * Write your own problem of this type, and then solve it yourself using this tactic in two different ways.