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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

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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

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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.

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Joined: **15 Jan 2018 **

Posts: **147**

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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

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Expert Reply

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.

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Re: If x and y are integers, and w = x2y + x + 3y, which of the
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29 Oct 2018, 09:34

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Carcass wrote:

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.

This is a great candidate for testing all possible cases

If x and y are integers, then there are 4 possible cases:

case 1) x is ODD and y is ODD

case 2) x is ODD and y is EVEN

case 3) x is EVEN and y is ODD

case 4) x is EVEN and y is EVEN

For each case, we can either apply the rules for even and odd numbers, or we can plug in nice values (0 for even and 1 for odd)

Let's use the second strategy.

case 1) x is ODD and y is ODD

w = x²y + x + 3y = (1²)(1) + 1 + 3(1) = 5

So, w is ODD

case 2) x is ODD and y is EVEN

w = x²y + x + 3y = (1²)(0) + 1 + 3(0) = 1

So, w is ODD

case 3) x is EVEN and y is ODD

w = x²y + x + 3y = (0²)(1) + 0 + 3(1) = 3

So, w is ODD

case 4) x is EVEN and y is EVEN

w = x²y + x + 3y = (0²)(0) + 0 + 3(0) = 0

So, w is EVEN

Check the statements:

A. If w is even, then x must be even.

w is even ONLY in case 4. In case 4, x is even

This statement is TRUE

B. If x is odd, then w must be odd.

x is odd in cases 1 and 2.

In both cases w is odd

This statement is TRUE

C. If y is odd, then w must be odd.

y is odd in cases 1 and 3.

In both cases w is odd

This statement is TRUE

D. If w is odd, then y must be odd.

w is odd in cases 1, 2 and 3

In case 2, y is EVEN

This statement is NOT true

Answer: A, B, C

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Re: If x and y are integers, and w = x2y + x + 3y, which of the
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17 Nov 2018, 04:51

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Is thEre a shorter way to get the answer?

Re: If x and y are integers, and w = x2y + x + 3y, which of the
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17 Nov 2018, 05:28

Expert Reply

ShefaliSahu wrote:

Is thEre a shorter way to get the answer?

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.

since all choices have w as odd or w as even, let us check when can w be odd and when even

1) when both x and y are even, w becomes \(E^2*E+E+3E=E\)

2) when both x and y are odd, w becomes \(O^2*O+O+3O=O\)

3) when both x and y are different, w becomes \(O^2*E+O+3E=E+O+E=O\)

so

w will be even, only if x and y are even

w will be odd, whenever any one or both of x and y are odd

let us see choices..

A. If w is even, then x must be even. => w will be even, only if x and y are even YES

B. If x is odd, then w must be odd. =>w will be odd, whenever any one or both of x and y are odd YES

C. If y is odd, then w must be odd. =>w will be odd, whenever any one or both of x and y are odd YES

D. If w is odd, then y must be odd. =>w will be odd, whenever any one or both of x and y are odd so it may be that just one , that is x , is odd and y is not.. so, not MUST

A, B, and C

Re: If x and y are integers, and w = x^2y + x + 3y, which
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15 Aug 2024, 19:01

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Re: If x and y are integers, and w = x^2y + x + 3y, which [#permalink]

15 Aug 2024, 19:01
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