If x and y are integers and xy^2 is a positive odd integer, which of

If x and y are integers and xy^2 is a positive odd integer, which of the following must be true?
I. xy is positive.
II. xy is odd.
III. x + y is even.

A. I only
B. II only
C. III only
D. I and II
E. II and III

As \(x\) and \(y\) are integers and \(xy^2\) is a positive odd integer then both \(x\) and \(y\) must be odd.

Now, let's consider each option:
I. xy is positive --> not necessarily true as \(x\) can be positive odd number and \(y\) can be negative odd number.

II. xy is odd --> as both \(x\) and \(y\) are odd then \(xy=odd*odd=odd\), hence this statement is always true.

III. x + y is even --> as both \(x\) and \(y\) are odd then \(x+y=odd+odd=even\), hence this statement is always true.

Answer: E (II and III)

As for your example: if \(x=3\) and \(y=-3\) then \(x+y=3-3=0=even\), so statemet III is still satisfied.

Hope it's clear.

But Bunuel I guess 0 is neither odd nor even number. Thats why I ruled out option III.Let me know your thoughts.

Zero is an even integer.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.
An even number is an integer of the form \(n=2k\), where \(k\) is an integer.

So for \(k=0\) --> \(n=2*0=0\).

For more on number properties check: math-number-theory-88376.html

Hope it's clear.

xy^2 is a positive odd integer

The moment you come across such information, first of all, think what it implies in this question... If you do, getting to your answer will be quick and easy... Given that x and y are integers,

xy^2 is positive implies that x is positive (Since y^2 is never negative so pos = pos*pos). y is either positive or negative (neither of them is 0)

xy^2 is odd means both x and y are odd. If either one of them were even, xy^2 would have been even.

So we get the following: x - positive odd; y - odd

Now run through the statements to get your answer.
Ⅰ. xy is positive. - Not necessary. If y is negative, xy is negative
Ⅱ. xy is odd. - Necessary since both x and y are odd.
Ⅲ. x + y is even. - Odd + Odd = Even hence necessary

Answer (E)

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