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If x and y are integers, is xy even?
1) x = y + 1
2)
x/y is an even integer

a)Exactly one of the statements can answer the question
b)Both statements are required to answer the question
c)Each statement can answer the question individually
d)More information is required as the information provided is insufficient to answer the question
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If x and y are integers, is xy even?1) x = y + 12)... more x/y is an e...
If either x or y is zero, then xy=0=even, because zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

 
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Most Upvoted Answer
If x and y are integers, is xy even?1) x = y + 12)... more x/y is an e...
Statement 1: x = y

If x = y, then we can substitute y for x in the expression xy, which gives us y * y = y^2.
Since y^2 is an integer, we can conclude that xy is even.

Statement 2: x/y is an even integer

If x/y is an even integer, it means that the ratio x/y can be expressed as a fraction where both x and y are integers, and the numerator x is divisible by the denominator y.
In other words, y is a factor of x.

To determine if xy is even, we need to consider two scenarios:
1. If y is an even number, then x must also be even (since y is a factor of x). In this case, xy is even.
2. If y is an odd number, then x can be either even or odd. If x is even, then xy is even. However, if x is odd, then xy is odd.

Therefore, statement 2 alone is not sufficient to determine if xy is even.

Conclusion:
From statement 1, we know that xy is even if x = y.
From statement 2, we cannot determine if xy is even.

Since statement 1 alone is sufficient to answer the question, but statement 2 alone is not, the correct answer is option C) Each statement can answer the question individually.
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