2018/011) Let $P(n) = (n+1)(n+(n+3)(n+5)(n+7)(n+9)$ What is the largest integer that is a divisor of P(n) for n even

Choose 2 values quite far off (as, if we chose 2 consecutive values then 4 terms shall be same
We have $P(2) = 3 * 5 * 7 * 9 *11$
and $P(10) = 11 * 13 * 15 * 17 * 19$
$GCD(P(2),P(10) = 15$
So GCD of the numbers shall be factor of 15
Now we have $(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)(n+3)(n+2)(n+1)(n+3) \pmod 3$
and it is product of 3 consecutive numbers we have this is divisible by 3
Now we have $(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)(n+3)(n+5)(n+2)(n+4) \pmod 5$
and it is product of 5 consecutive numbers we have this is divisible by 5
So it is divisible by 15
So 15 is the required number