Q. Let X ,Y and Z be distinct integers where x and y are odd and positive ,and Z is even and positive. Which one of the following statement can't be true ?
A. (x-z)square is even
B. (x-z)ysquare is odd
C. (x-z)y is odd
D. (x-y)squarez is even
Please tell me the correct answer of this question and please explain it in detail and I want the full solution
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NOTE::
§ Odd multiplied by odd is always odd.
§ odd - even or even - odd is always odd
§ even - even is always even
§ square of even term is even
§ square of odd term is odd.
§ odd - odd is even
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a) is the answer!
_________________________
Since , @ d, x and y are odd so their differences will be an even term whose square is even and even multiplying with z will also give even term so statement of #d is true.
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@c, x- z will be an odd term whose multiplication with y (odd) will give an odd term. So, statement #c is true.
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@b, according to c (x- z)y is odd so it's square is also odd. So statement #b is true.
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@ a, (x- z) is odd and it's square is also odd.
But the statement states that it's even. So, statement #a is false.
_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_
Hope this is ur required answer
Proud to help you
§ Odd multiplied by odd is always odd.
§ odd - even or even - odd is always odd
§ even - even is always even
§ square of even term is even
§ square of odd term is odd.
§ odd - odd is even
----------
a) is the answer!
_________________________
Since , @ d, x and y are odd so their differences will be an even term whose square is even and even multiplying with z will also give even term so statement of #d is true.
__ __ __ __ __ __ __ __ __ __ __
@c, x- z will be an odd term whose multiplication with y (odd) will give an odd term. So, statement #c is true.
__ __ __ __ __ __ __ __ __ __ __
@b, according to c (x- z)y is odd so it's square is also odd. So statement #b is true.
__ __ __ __ __ __ __ __ __ __ __
@ a, (x- z) is odd and it's square is also odd.
But the statement states that it's even. So, statement #a is false.
_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_
Hope this is ur required answer
Proud to help you
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