If k is an integer, is k even?

Question : Is k even?

Statement 1: k + 5 is not an even integer

We must know that
Even + Even = Even
Even + Odd = Odd
Odd + Odd = Even

k+odd(5) = Odd
hence K must be even
SUFFICIENT

Statement 2: k/2 is an even integer

i.e. k = 2 x Even
i.e. k = Even
SUFFICIENT

Answer: Option

Here we go:

The key to remember is

{odd + odd} = even ----(1)
{even + odd} = odd ----(2)
{even + even} = even ----(3)

St1: k + 5 is not an even integer

5 is an odd integer
Equation (2) holds true here

St1 is sufficient

St2: k/2 is an even integer

k/2 = even integer

k = 2 * even integer =====> even integer

St2 is also sufficient.

Option D is correct

I tried this by inputting a few numbers.

(1) -1+5 = even
0+5 = odd
did this for a couple numbers and only the even values of K yield a "not even" number. So 1 is sufficient because I can definitively answer the question.

(2) I did the same thing here, and again I can definitively answer the question. So also sufficient.

Answer I came to is D.

(1) k+5 is not even
here k may be 0 as 0 is considered as an integer. so k is neither even nor odd.
also k can be 2, in which case k is even
so (1) is NSF
(2) as k/2 is even, K has to be even. hence SF

Answer is B

But isn't 0 an even integer?

Hi kler515,

You've made a small mistake with how you dealt with Fact 1.

There are a number of facts that you need to know about the number 0 (and you'll likely be tested on ALL of them on the Official GMAT):

1) 0 is an integer
2) 0 is even
3) 0 is called a "null value" (it's neither positive, nor negative)
4) 0 is a multiple of everything

So, since 0 is even, and all of the other examples that you could come up with are ALSO even, Fact 1 is actually sufficient.

GMAT assassins aren't born, they're made,
Rich

Since k is an integer, it can take any value from -infinity to +infinity including 0.
We can obtain an odd number only when an ODD number is ADDED to an EVEN number. In all other cases, we always get an even number (i.e. Even + Even, Odd + Odd)

1) Using first condition, we know that k + 5 is an ODD integer.
Hence we're sure that k is an EVEN integer because 5 is odd.
Hence this alone can help us answer the question.

2) Using second condition, we know that k/2 is an even integer.
The basic rule to test Even/Odd numbers is that if any integer when divided by 2 given an even integer then that integer is even.
Since k/2 is an even integer so k is an even integer.
Hence this alone can help us answer the question.

Hence the answer is D.

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VERITAS PREP OFFICIAL SOLUTION

Correct Answer: D

Statement (1) indicates that k plus an odd number is not even. Two odd numbers would add up to an even number, so we know that k cannot be odd because k + 5 is not even. Therefore, k is an even number and statement (1) is sufficient.

Statement (2) indicates that k/2 is an even integer. In order to get an even integer when dividing an integer by 2, the original integer must be a multiple of 4 (i.e., 8/2 = 4 or 20/2 = 10). Thus k is an even number and statement (2) is sufficient as well. Since each statement alone is sufficient to answer the question, the correct response is D.
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If k is an integer, is k even?

(1) k + 5 is not an even integer
(2) k/2 is an even integer


State(1) - k+5 = odd
k+odd = odd
k is even (sufficient)

State(2) - k/2 = even
k = even (sufficient)

Both alone are sufficient.

If k is an integer, is k even?

1. k + 5 is not an even integer

2. k /2 is an even integer

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked
C. Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient
D. EACH statement ALONE is sufficient to answer the question asked
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed

I think answer is B , but official answer is different . Please help how k+5 can always be even as k could be 0.

1) k+5 is not an even integer => k+5 is odd, so k is even. Sufficient
2) k/2 is even. => k is even. Sufficient

Hence, D

Merging topics. Please refer to the discussion above.
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Hi ,
Bunel , please help me understand how statement 1 is correct as k could be 0 then K+5 will be odd and k will not be even.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. 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.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: number-properties-tips-and-hints-174996.html
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(1) k + 5 is not an even integer
k+odd= odd

k has to be even. Sufficient.

(2) k/2 is an even integer
K/2 is even integer means that k is in the form 4n

sufficient.

D is the answer
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