Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.

1)

O x E = E + E = E Thus, K = even
O x O = O + E = O K = odd does not satisfy the statement. Thus, k must be even = sufficient

2)

O x O = O + O = E Thus, K = odd does not satisfy statement
E x E = E + O = O K = even does satisfy statement, thus k must be even = sufficient

= D imo

k^2 + 1 = odd

so k is even

what if k = \(\sqrt{2}\)

then we have k^2 + 1 = 3 as k =\(\sqrt{2}\)

but here k is \(\sqrt{2}\) , which is not an integer

so b wouldn't be sufficient imo

This is a tricky one. The question asks if \(k\) is an even integer.

Statement 1
If \(k = 2\), then the answer to the question is YES.
If \(k = \frac{2}{5}\), then the answer to the question is NO.
Not sufficient.

Statement 2
If \(k = 2\), then the answer to the question is YES.
If \(k = \sqrt{2}\), then the answer to the question is NO.
Not sufficient.

Combine Statements 1 & 2
There has to be a simpler approach (please show me). Here's what I did...

Statement 1 in equation form is:
\(5k + 2 = 2n\), where \(n\) is an integer.
Solve for k,

\(5k + 2 = 2n\)
\(5k = 2n - 2\)

\(k = \frac{2(n - 1)}{5}\) .........Equation 1

Statement 2 in equation form is:
\(k^2 + 1 = 2m + 1\), where \(m\) is an integer.
Solve for k,

\(k^2 + 1 = 2m + 1\)
\(k^2 = 2m\)

\(k = \sqrt{2m}\) .........Equation 2

Equate the two equations and simplify:
\(\frac{2(n - 1)}{5} = \sqrt{2m}\)

\(\frac{4(n - 1)^2}{25} = 2m\)

\(4(n - 1)^2 = 50m\)
\(2(n - 1)^2 = 25m\) .........Equation 3

Since \(n\) and \(m\) are integers, the LHS and RHS are integers. Also, the RHS is divisible by 25, so the LHS must be divisible by 25. Thus \((n - 1)^2\) must be divisible by 25. This means
\((n - 1)^2 = 25r\), where \(r\) is an integer.
\(n - 1 = 5\sqrt{r}\).
\(n - 1\) is an integer and thus \(5\sqrt{r}\) is an integer. And since \(r\) is an integer, \(\sqrt{r}\) must be an integer. Thus, \(n - 1\) is a multiple of 5. In other words,
\(n - 1 = 5s\) for some integer \(s\). .........Equation 4

Now we know that \(n\) can't be every single integer, but a subset of integers, e.g. {6,11,16,21,etc}. Let's substitute Equation 4 into Equation 1 and see what we get for \(k\):

\(k = \frac{2(n - 1)}{5} = \frac{2(5s)}{5} = 2s\)

This tells us that \(k\) must be an even integer. Both statements combined are sufficient.

Answer: C

Wow I think I just figured out the simpler approach...

Statement 1 says that 5k + 2 is even. Or more specifically (redundantly?), an even integer. So 5k must be an even integer. k can either be a fraction with 5 in the denominator and an even number in the numerator, or k can be an even integer.
Statement 2 says that k^2 + 1 is odd.... i.e. an odd integer. So k^2 must be an even integer. k can either be the square root of an even integer, or k can be an even integer.
Combining the 2 statements, the one thing in common is that k is an even integer.

Answer: C

Statement I:

Consider \(K =\frac{2}{5}, 2\). Not Sufficient.

Statement II:

Consider \(K = \sqrt{2}, 2\). Not Sufficient.

Combining I & II:

K has to be even number.

Under root of a number can never be equal to a fraction of two integers.

Eg. √2 cannot be expressed as a/b where a & b are Integers.

Hence, when evaluating St.1+2:
Only option left for 'k' is to be an integer.

It is already clear from each statement that if k is an integer, then it's an even Integer.

So, imo, correct Ans is C

Posted from my mobile device

Question : Is k even?

St 1: \(5k + 2\) even

If k = 2, answer to the QUESTION is YES

If k = 2/5, answer to the QUESTION is NO (Insufficient)

St 2: \(k^2 + 1\) is odd

If k = 2, answer to the QUESTION is YES

If k = \(\sqrt{2}\) or\(\sqrt{6}\), answer to the QUESTION is NO

Combining we get k = even

Hence (C)

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

Hello guys,

For the combination, what if k = 0 ? (Both statements are satisfied but it is not even)
0 is considered as non-even and non-odd, right?

Thanks

Posted from my mobile device

Nope 0 is considered to be even (as it's divisible by 2). You may be thinking of the rule about positives and negatives: 0 is neither positive nor negative.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.