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Re: Numbers again............ [#permalink]
27 Aug 2009, 11:01

xy^2 is a positive odd integer

1) Only way for this to hold true is that if x and \(y^2\) are both odd. As ExO and ExE are even

2) Further Y must be odd. The only way \(y^2\) to be odd is OxO.

Therefore we know both X and Y are odd

i) Not inferrable since we are squaring Y which eliminates +/- inference ii)As noted in premise 1 OxO is always odd so true iii) For addition the number property rule is this O+O = E E+E = E and O+E = O Therefore always even and true

Re: Numbers again............ [#permalink]
17 Jan 2010, 18:52

The answer is E.

I.- xy is positive. It's not possible to determine whether xy^2 could be -x*-y (i.e -3*-9 = +27) or x*y (i.e 3*9 = +27). II.- xy is odd. Anytime an ODD number is multiplied by an EVEN number the answer will be an EVEN number. Since xy^2 is ODD it implies that xy is ODD (.i.e. 27^2 =729 an ODD number). III.- x + y is even. Always an ODD number plus another ODD number is equal to a EVEN number (.i.e. 3 + 9 =12).

Re: Numbers again............ [#permalink]
20 Dec 2010, 11:35

Expert's post

Eshika wrote:

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.

Re: Numbers again............ [#permalink]
20 Dec 2010, 11:42

Bunuel wrote:

Eshika wrote:

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.

Re: Numbers again............ [#permalink]
21 Dec 2010, 13:22

Expert's post

milind1979 wrote:

16. 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

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

Re: If x and y are integers and xy^2 is a positive odd integer [#permalink]
15 Jan 2012, 21:14

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

xy^2 is odd implies both x and y are Odd

Ⅰ. xy is positive -> y can be positive or negative Ⅱ. xy is odd -> true, can be positive or negative depending on y. Ⅲ. x + y is even -> true.

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