If $n$ is any integer, which of the following is an odd number?
(a) $n^{2}-n+2$
(b) $n^{2}+n+2$
(c) $n^{2}+n+5$
(d) $n^{2}+5$
(e) $n^{3}+5$
The correct option is (c).

Let's recall that the sum or difference of numbers of the same parity is an even number:

$$
\text{even} \pm \text{even} = \text{even} \quad \text{and} \quad \text{odd} \pm \text{odd} = \text{even}.
$$

Observe that $n^{2}$ and $n^{3}$ can be even or odd, so $n^{2}+5$ and $n^{3}+5$ can be odd or even, depending on whether $n$ is even or odd. The remaining options are (a), (b), and (c).

Solution 1: Both $n^{2}-n$ and $n^{2}+n$ are the sum and difference of two numbers that always have the same parity, so these numbers will always be even, just like $n^{2}-n+2$ and $n^{2}+n+2$. Finally, the correct option is (c), because $n^{2}+n+5=\left(n^{2}+n\right)+5$, which is the sum of an even and an odd number, will always be an odd number, for any integer value of $n$.

Solution 2: Observe that $n^{2}-n=n(n-1)$ and $n^{2}+n=n(n+1)$ are the product of two consecutive numbers, so they are always even, just like $n^{2}-n+2$ and $n^{2}+n+2$. Finally, the correct option is (c), because $n^{2}+n+5=\left(n^{2}+n\right)+5$ is the sum of an even number and an odd number, which is always odd, for any integer value of $n$.