Largest divisor of $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$ 
Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?

I can see that obviously it is divisible by $3$ and $5$ and hence answer is $15$ but I can't prove it.

$$P(10)=(11)(13)(15)(17)(19)=3.5.11.13.17.19$$
$$P(20)=(21)(23)(25)(27)(29)=3^4.5^2.7.23.29$$
we can see that
 $$gcd(P(10),P(20))=15$$
$$P(n)=(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)n(n+2)(n+1)n \equiv 0 \mod 3$$
since $n(n+1)(n+2) \equiv 0 \mod 3.$, that is product of 3 consecutive numbers must be divisible by $3$.
Also, 
$$P(n)=(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)(n+3)n(n+2)(n+4)\equiv 0 \mod 5$$
since product of 5 consecutive numbers is divisible by $5$.
Hence $15$ is the largest number that divides $P(n)$ for all even $n$.
