To an integer, we add one-third of the integer, half of the integer's square, and one-sixth of the integer's cube. Prove that we always get an integer.

We start with an integer $n$. We need to show that the expression
$$\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$$
is always an integer.

First, let's find a common denominator for the fractions. The least common multiple of 3, 2, and 6 is 6. We rewrite the expression with a common denominator:
$$\frac{2n}{6} + \frac{3n^2}{6} + \frac{n^3}{6} = \frac{2n + 3n^2 + n^3}{6}.$$

Next, we need to show that the numerator $2n + 3n^2 + n^3$ is always divisible by 6. We can factor the numerator:
$$2n + 3n^2 + n^3 = n(2 + 3n + n^2).$$

We need to show that $n(2 + 3n + n^2)$ is divisible by 6. This means it must be divisible by both 2 and 3.

1. Divisibility by 2:
- If $n$ is even, then $n$ is divisible by 2, so $n(2 + 3n + n^2)$ is divisible by 2.
- If $n$ is odd, then $2 + 3n + n^2$ is even because:
- $2$ is even.
- $3n$ is odd (since $n$ is odd).
- $n^2$ is odd (since $n$ is odd).
- The sum of an even number and two odd numbers is even.
- Therefore, $n(2 + 3n + n^2)$ is divisible by 2.

2. Divisibility by 3:
- If $n$ is divisible by 3, then $n$ is divisible by 3, so $n(2 + 3n + n^2)$ is divisible by 3.
- If $n$ is not divisible by 3, then $n$ can be either $1$ or $2$ modulo 3.
- If $n \equiv 1 \pmod{3}$, then:
$$2 + 3n + n^2 \equiv 2 + 3 \cdot 1 + 1^2 \equiv 2 + 3 + 1 \equiv 6 \equiv 0 \pmod{3}.$$
- If $n \equiv 2 \pmod{3}$, then:
$$2 + 3n + n^2 \equiv 2 + 3 \cdot 2 + 2^2 \equiv 2 + 6 + 4 \equiv 12 \equiv 0 \pmod{3}.$$
- Therefore, $n(2 + 3n + n^2)$ is divisible by 3.

Since $n(2 + 3n + n^2)$ is divisible by both 2 and 3, it is divisible by 6. Therefore, the expression
$$\frac{2n + 3n^2 + n^3}{6}$$
is always an integer. This completes the proof.