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10 Jul 2006, 07:53
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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) ii and iii
(E) i and ii
Appreciate if someone can help me solve this with an explanation.
i. xy is positive
ii. xy is odd
iii. x+y is even
(A) i only
(B) ii only
(C) iii only
(D) ii and iii
(E) i and ii
Appreciate if someone can help me solve this with an explanation.
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i. wrong, counterexample: x=1 y=-1. [xy^2=1, xy=-1]
ii. correct
Since xy^2 is odd, we know that x and y^2 must be odd. Since y^2=y*y is odd, y must be odd, too. => x*y is odd.
iii. correct
Since both x and y are odd and can be written as 2k+1 respectively 2n+1 (k, n integers)
Their sum is 2k+1+2n+1 = 2k+2n+2. This is divisible by 2. Hence x+y is even. Note that 2k+2n+2=0 is possible (but 0 is even, too).
Therefore, D is correct.
ii. correct
Since xy^2 is odd, we know that x and y^2 must be odd. Since y^2=y*y is odd, y must be odd, too. => x*y is odd.
iii. correct
Since both x and y are odd and can be written as 2k+1 respectively 2n+1 (k, n integers)
Their sum is 2k+1+2n+1 = 2k+2n+2. This is divisible by 2. Hence x+y is even. Note that 2k+2n+2=0 is possible (but 0 is even, too).
Therefore, D is correct.
10 Jul 2006, 10:08
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(D) ii and iii
Given (xy)^2 is +ve odd.. i.e. xy has to be odd => x & y are odd (as prod of odd x even would be even).
(i) cannot be true as x or y could be -ve.
Therefore D. ii & iii
Given (xy)^2 is +ve odd.. i.e. xy has to be odd => x & y are odd (as prod of odd x even would be even).
(i) cannot be true as x or y could be -ve.
Therefore D. ii & iii
