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Testing cases for even and odd numbers. In the last video we talked about the properties of even and odd numbers and we’re gonna use those properties here. So if you didn’t see that video, I’d highly recommend watching that before watching this video. So, here we have a practice problem.
I’ll suggest pause the video right here and do this practice problem, and then we’ll talk about it. If P and Q are integers and if P squared plus PQ is an odd number, what do we know about P and Q? The easiest way to approach this is to plug in numbers for different cases.
We can simply use 1 for our odd number and 2 for our even number. And notice we’re gonna have four different cases, four different possibilities here. They both could be even, both P and Q could be even. Both P and Q could be odd numbers. Or it may be that P is even and Q is odd. Orr it may be that P is odd and Q is even.
So, we have to test all four of those cases separately. Case one, P and Q are both even. So, we’re just gonna plug in 2, P equals 2 and Q equals 2. We are gonna get, 2 squared, plus 2 times-2, 4 plus 4 is 8. That’s an even number, that doesn’t work. We’ll try both odd.
Pretend that they are both equal to 1, P equals 1, Q equals 1. We plug in, we have 1 plus 1 is 2, which is an even number. That doesn’t work. Case three, P is even and Q is odd, so now we’re going to plug in P equals 2 and Q equals 1. And here we’re gonna have 4 plus 2 is 6, that doesn’t work.
That’s even. So then the final case, we’re wondering now, is it possible for this to be odd? The final case, we’ll try P as odd and Q as even. Now we’re gonna plug in P equals 1 and Q equals 2, and here we’re gonna get 1 plus 2 is 3. That definitely is an odd number.
That works and that’s the only case among the four that works. Therefore we know, in this problem it must be true that P is an odd number and Q is an even number. We are able to figure that out from the four-case method. We might also solve this problem using logic. Think about it this way.
Notice if P is even, it automatically makes both terms even. And then we get even plus even, which equals even and that doesn’t work. There’s no way we can get an odd number from even plus even. So therefore, P has to be odd. Now if P is odd, the first term is definitely odd. If the sum is odd, the second term must be even.
If P is odd, the only way P times Q can be even, is if Q is an even number. Therefore, P is odd and Q is even. So, that’s the way to solve it with logic. Ultimately, solving it with logic is more direct, it’s quicker, but it can be a bit harder to learn how to think that way. And the four-method approach is a little more methodical, that can be good when you’re getting kind of warmed up to this sort of problem.
Here’s another one of these problems. So again, pause the video, and then we’ll talk about how to solve this. If P and Q are integers, and the quantity 4P plus Q is an odd number, what must be true? Let’s start with the four-case method.
If they’re both even, so we’re gonna plug in P equals 2 and Q equals 2, we get 8 plus 2, which is 10, which is an even number. Now we’ll try both odd. We’ll plug in 1 and 1. And we get 4 plus 1, that’s 5. That definitely works.
So case one doesn’t work, case two works. Case three, P is even and Q is odd. So we’re, now we’re gonna plug in P equals 2 and Q equals 1. And we get 8 plus 1, 9, that’s an odd number, that works. So, we have two cases that work. Now we’ll try case four, P is odd, Q is even.
We get 4 plus 2 is 6, that doesn’t work. So what’s, exactly is the pattern here? Notice that the cases that work are the cases where Q is odd. If Q is odd, then the expression is odd, and if Q is even, then the expression doesn’t work. So we, the conclusion that we can draw definitely is, Q is odd, and we really can’t draw anything about P.
It’s definitely true that Q is odd, and we have, we can draw no conclusion about P. Now again using logic, think about this. 4P must be even, because 4 is an even number. So, the only way that the sum will be odd is if Q is an odd number. So, Q must be odd.
4P will be even regardless of whether P is even or odd, so whether P is even or odd does not matter at all to the result. If you are still developing your number sense, then you may find it helpful to start with the four-case method of plugging in numbers. Even if you solve the problem that way, it’s a good idea to go back and see if you can solve the problem with logic as well.
Often, a logical solution is shown in the official explanations to a problem. And again, practicing the logical solution ultimately will be a much deeper form of mathematical understanding. Applying the four cases is kind of rote, that’s not real deep mathematical thinking. Ultimately what you want to do is practice your mathematical thinking.
In summary. In this video we discuss a four-case method of plugging in numbers to solve even and odd questions, and we dis, discussed logical approaches to such problems as well.
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