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If x and y are integers, and w = x^2y + x + 3y, which
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20 Nov 2017, 10:47
20 Nov 2017, 10:47
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If x and y are integers, and \(w = x^2y + x + 3y\), which of the following statements must be true?
Indicate all such statements.
A. If w is even, then x must be even.
B. If x is odd, then w must be odd.
C. If y is odd, then w must be odd.
D. If w is odd, then y must be odd.
Kudos for correct solution.
Indicate all such statements.
A. If w is even, then x must be even.
B. If x is odd, then w must be odd.
C. If y is odd, then w must be odd.
D. If w is odd, then y must be odd.
Kudos for correct solution.
Show: :: OA
A, B, and C
If x and y are integers, and w = x2y + x + 3y, which of the
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27 Oct 2018, 16:39
27 Oct 2018, 16:39
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If x and y are integers, and \(w = x^2y + x + 3y\), which of the following statements must be true?
Indicate all such statements.
A. If w is even, then x must be even.
B. If x is odd, then w must be odd.
C. If y is odd, then w must be odd.
D. If w is odd, then y must be odd.
Indicate all such statements.
A. If w is even, then x must be even.
B. If x is odd, then w must be odd.
C. If y is odd, then w must be odd.
D. If w is odd, then y must be odd.
General Discussion
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Re: If x and y are integers, and w = x^2y + x + 3y, which
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21 Feb 2018, 10:56
21 Feb 2018, 10:56
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A good technique for problems asking what must be true is to try and make them not true. If you can't, it must be true. Also, a good technique for odd/even questions is to just test using your own odd and even numbers. Since every odd number will behave the same as every other in regards to oddness or evenness, it doesn't matter what number you choose. I like to use 0 and 1 for even and odd, respectively, since they tend to make the math very simple.
Also, it seems like a good idea to factor the equation since they probably didn't give it to us in the easiest format. One way to factor it is
y(3 + x^2) + x
A: I'll try to find a way to make w even when x is odd. Plugging in 1 for x gets me 4y + 1. It doesn't matter what y is; what we've got here is an even number plus 1, which must be odd. So there's no way to make an even number if x is odd. Thus A is in.
B: Well we just tried making x odd and found that w must be odd if x is, so B is in.
C: Let's see what happens if we substitute y for 1: We'll get 1(3 + x^2) + x or just x^2 + x + 3. We'll need to check what happens when x is odd and when x is even:
If x is 0, we get 0 + 0 + 3, which is odd.
If x is 1, we get 1 + 1 + 3, which is odd.
So it looks like C is in as well.
D: Let's try making y even and see whether we can get w to be odd:
If we make y = 0, we'll have 0(3 + x^2) + x, which is just x. So if we pick x to be odd, then w would be odd. And since we got an odd w with an even y, D is out.
Also, it seems like a good idea to factor the equation since they probably didn't give it to us in the easiest format. One way to factor it is
y(3 + x^2) + x
A: I'll try to find a way to make w even when x is odd. Plugging in 1 for x gets me 4y + 1. It doesn't matter what y is; what we've got here is an even number plus 1, which must be odd. So there's no way to make an even number if x is odd. Thus A is in.
B: Well we just tried making x odd and found that w must be odd if x is, so B is in.
C: Let's see what happens if we substitute y for 1: We'll get 1(3 + x^2) + x or just x^2 + x + 3. We'll need to check what happens when x is odd and when x is even:
If x is 0, we get 0 + 0 + 3, which is odd.
If x is 1, we get 1 + 1 + 3, which is odd.
So it looks like C is in as well.
D: Let's try making y even and see whether we can get w to be odd:
If we make y = 0, we'll have 0(3 + x^2) + x, which is just x. So if we pick x to be odd, then w would be odd. And since we got an odd w with an even y, D is out.
