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

integer.

Statement 1If \(k = 2\), then the answer to the question is YES.

If \(k = \frac{2}{5}\), then the answer to the question is NO.

Not sufficient.

Statement 2If \(k = 2\), 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 & 2There 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 3Since \(n\) 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 4Now 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