By which number?

If $n\equiv 0\bmod 3$ then $n(2n^2+7)\equiv 0\bmod 3$ . If $n\equiv 1\bmod 3$ or $n\equiv 2\bmod 3$ then $n^2\equiv 1\bmod 3$ . So $\begin{aligned} n(2n^2+7) &\equiv n(2+7)\bmod 3\\ &\equiv 0\bmod 3 \end{aligned}$ Therefore $n$ is always divisible by $\boxed{3}$ .

We note that $f(n) = n(2n^2 + 7)$ is divisible by 3 for the first few $n$ 's. Let us prove that $3 \mid f(n)$ for all natural number $n$ as follows.

All natural number $n$ can be represented as $n = \begin{cases} 3m & \text{if } n \bmod 3 = 0 \\ 3m + 1 & \text{if } n \bmod 3 = 1 \\ 3m -1 & \text{if } n \bmod 3 = 2 \end{cases}$ , where $m$ is a natural number.

If $n \bmod 3 = 0$ , then $f(3m) = 3m\left(2(3m)^2 + 7\right) \equiv 0 \text{ (mod 3)}$ .

If $n \bmod 3 = 1$ , then

$\begin{aligned} f(3m+1) & \equiv (3m+1)\left(2(3m+1)^2 + 7\right) \text{ (mod 3)} \\ & \equiv (1)\left(2(1)^2 + 7 \right) \text{ (mod 3)} \\ & \equiv 9 \equiv 0 \text{ (mod 3)} \end{aligned}$

If $n \bmod 3 = 2$ , then

$\begin{aligned} f(3m-1) & \equiv (3m-1)\left(2(3m-1)^2 + 7\right) \text{ (mod 3)} \\ & \equiv (-1)\left(2(-1)^2 + 7 \right) \text{ (mod 3)} \\ & \equiv -9 \equiv 0 \text{ (mod 3)} \end{aligned}$

Therefore, $\boxed 3 \mid n \left(2n^2+7\right)$ for all natural numbers $n$ .

Answer is 3.

Use Euclid division lemma by using (a=bq+r),b=3