Irish Mathematical Olympiad

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 $2310$.