Finding n?

Notice that $\displaystyle \overline{n10} \leq 910$ since it is a three digit number. Also notice that $\displaystyle \overline{n10} > n^4$ since it equals $n \cdot (n + 1) \cdot (n + 3) \cdot (n + 5)$ .

Thus,

$\displaystyle \begin{aligned} n^4 & \leq 910 \\ n & \leq \sqrt[4]{910} \\ n & \leq 5.492 & \small \color{#3D99F6} n \in \mathbb{N} \\ n & \leq 5 \end{aligned}$

The only prime numbers $n \leq 5$ are $2$ , $3$ , and $5$ . Recall that if $n$ is prime, $n + 1$ must also be prime, since it is one of the prime factors of $\displaystyle \overline{n10}$ . Clearly, $n \neq 3$ and $n \neq 5$ since $4$ and $6$ are not prime.

Thus, $\displaystyle \boxed{n = 2}$ and the product is $210$ . We see that $2, 3, 5, 7$ are all prime and $2 \cdot 3 \cdot 5 \cdot 7 = 210$ .

Since n and n+1 are both consecutive primes, 1 of them is even, so 2 is 1 of the prime factors. If n+1 would be 2(so n=1) the n+2 is 4, which is not prime. Hence n=2.

Other than the first prime 2, all larger primes are odd. If $n+1$ , $n+3$ , and $n+5$ are primes, they must be odd. For three of them to be odd, $n$ must be even. Since $n$ is a prime and even, it must be 2 and the four primes are 2, 3, 5, and 7; and their product is 210. Therefore $n=\boxed{2}$ .

$210= 2\times 3\times 5\times 7$ , thus $n=2$ .