Irska

Let $m$ be the g.c.d. of $E(1),E(2),E(3),\dots ,E(2009)$. Since $m|E(1)=2\cdot 3\dots 11$, it follows that any prime divisor of $m$ is less than or equal to $11$. Let $p$ be a prime number such that $p\nmid m$. Since $p\le 11<2009$, it follows that $m|E(p)=p(p+1)(2p+1)(3p+1)\cdots (10p+1)$. Observe that $p+1,2p+1,3p+1,\dots ,10p+1$ are relatively prime to $p$. so $E(p)$ (and thus $m$) is divisible by $p$ but not by $p^2$. We have thus proved that $m$ is not divisible by the square of any prime number.

Since $m|E(1)=2\cdot 3\cdots 11$, it follows that $m$ divides the product of all prime numbers less than or equal to $11$, that is, $m|2310$.

To show that $m=2310$ it is enough to prove that for all $n\ge 1$, the number $E(n)$ is divisible by $2310$.

Let $n\ge 1$. Then, one of the numbers $n$ or $n+1$ is divisible by $2$, so $2|E(n)$. Similarly, one of the numbers $n,n+1,2n+1$ is divisible by $3$ so $3|E(n)$. Then, one of the numbers $n,n+1,2n+1,3n+1,4n+1$ is divisible by $5$ which yields $5|E(n)$. In the same manner we obtain $7|E(n)$ and $11|E(n)$. Therefore $E(n)$ is a multiple of $2\cdot 3\cdot 5\cdot 7\cdot 11=2310$ and so, the g.c.d. is $2310$.