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Re: If x and y are integers, and w = x2y + x + 3y, which of the [#permalink]
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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.


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 [#permalink]
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Is thEre a shorter way to get the answer?
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Re: If x and y are integers, and w = x2y + x + 3y, which of the [#permalink]
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ShefaliSahu wrote:
Is thEre a shorter way to get the answer?


Quote:
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.


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