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28 Feb 2021, 21:16
Kudos
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Hello SME,
In below Question,
If X and Y are positive integer, Is Y an even Integer
St 1 X( X+2) + 1= XY
St 2 X ( X + Y) = Odd
I have doubt in infefrenc made for eliminating the St -1.
IF X - Odd
odd x odd + 1= odd x Y
even= Odd x Y
Y = Even/Odd ( This could be Fraction or Even but as per Question statement Y is Integer. Thus Y can be Even)
IF X -even
Even X Even +1= Even x Y
0dd= Even x Y
Y= Odd/Even ( This is fraction but Y cannot be fraction. Thus Inference is Y can be Odd or Even. Thus no definite Answer in both Case thus St-1 is not suffieint)
Please advise is any point missed in case when X - even.
In below Question,
If X and Y are positive integer, Is Y an even Integer
St 1 X( X+2) + 1= XY
St 2 X ( X + Y) = Odd
I have doubt in infefrenc made for eliminating the St -1.
IF X - Odd
odd x odd + 1= odd x Y
even= Odd x Y
Y = Even/Odd ( This could be Fraction or Even but as per Question statement Y is Integer. Thus Y can be Even)
IF X -even
Even X Even +1= Even x Y
0dd= Even x Y
Y= Odd/Even ( This is fraction but Y cannot be fraction. Thus Inference is Y can be Odd or Even. Thus no definite Answer in both Case thus St-1 is not suffieint)
Please advise is any point missed in case when X - even.
shameekv1989
01 Mar 2021, 03:00
Kudos
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vickytvn
When analyzing statement 1 what you are getting is that X has to be Odd and Y has to be even
Case 2 is invalid as you pointed out correctly that Y can not be a fraction. Thus only Case 1 i.e. if X is Odd Y is even which gives you the definite answer.
02 Mar 2021, 03:02
Kudos
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vickytvn
Statement 1 is sufficient because
if X is even condition will not exist, as when X is even both sides of the equation will not be equal.
i.e.
if X is even ,
then LHS is odd. and in no condition can RHS be odd, because RHS is XY and as X is even, no matter the value of Y (except fraction), it will always be even.
