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milind1979
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If x and y are integers and xy^2 is a positive odd integer, which of 01 Apr 2009, 03:25
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E
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73% (01:17) correct 27% (01:27) wrong based on 1498 sessions

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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
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E
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Bunuel
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If x and y are integers and xy^2 is a positive odd integer, which of Updated on: 30 Jun 2025, 05:57
Originally posted by Bunuel on 20 Dec 2010, 12:15.
Last edited by Bunuel on 30 Jun 2025, 05:57, edited 1 time in total.
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Eshika
I guess answer is B. III option is not always right.Reason:-

When x=3 and y=-3 xy^2=27(odd integer) but x+y=0 and that is not an odd integer.Hence answer will be B.

Let me have your inputs on the same.
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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.
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johnhillescobar
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If x and y are integers and xy^2 is a positive odd integer, which of 17 Jan 2010, 19:52
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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).
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If x and y are integers and xy^2 is a positive odd integer, which of 20 Dec 2010, 12:01
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I guess answer is B. III option is not always right.Reason:-

When x=3 and y=-3 xy^2=27(odd integer) but x+y=0 and that is not an odd integer.Hence answer will be B.

Let me have your inputs on the same.
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If x and y are integers and xy^2 is a positive odd integer, which of 20 Dec 2010, 12:31
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But Bunuel I guess 0 is neither odd nor even number. Thats why I ruled out option III.Let me know your thoughts.
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If x and y are integers and xy^2 is a positive odd integer, which of 20 Dec 2010, 12:35
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Eshika
But Bunuel I guess 0 is neither odd nor even number. Thats why I ruled out option III.Let me know your thoughts.
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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.
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If x and y are integers and xy^2 is a positive odd integer, which of 21 Dec 2010, 14:22
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milind1979
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
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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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If x and y are integers and xy^2 is a positive odd integer, which of 30 Jun 2025, 05:07
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For x+y as even(it is not necessary, it needs to be even), if we keep x as 3 and y as 8 which sums up as 11, but it also made 11x11 = 121 positive and odd, and x+y is also not even! any take regarding it? I'm unable to understand.
johnhillescobar
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).
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If x and y are integers and xy^2 is a positive odd integer, which of 30 Jun 2025, 05:59
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vikalpjain91
For x+y as even(it is not necessary, it needs to be even), if we keep x as 3 and y as 8 which sums up as 11, but it also made 11x11 = 121 positive and odd, and x+y is also not even! any take regarding it? I'm unable to understand.
Bunuel

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.

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Your confusion seems to come from the example you used: x = 3 and y = 8. But in that case, y = 8 is even, so xy^2 = 3 * 64 = 192, which is even, not odd. That example violates the condition given in the question that xy^2 is a positive odd integer.

Please refer to the complete solution in the quoted text.
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