Find the GCD of $10$ integers 
Find the greatest common divisor of the following ten integers: 
  \begin{align}
& 2000^3+3\cdot2000^2+2\cdot2000,\  2001^3+3\cdot2001^2+2\cdot2001,\  \dots, \\[5pt]
& \dots, \  2008^3+3\cdot2008^2+2\cdot2008,\  2009^3+3\cdot2009^2+2\cdot2009.
\end{align}

How are these numbers related to each other? Is there a formula for this?
 A: You've got $n(n+1)(n+2)$ for $n=2000,\ldots,2009$. You want divisors that they all share in common.
Given three consecutive integers, exactly one of them is divisible by $3$; therefore their product is divisible by $3$. That is true of all $10$ of the products, so $3$ is a common divisor. Among the ten products, there are some, such as $2000\times2001\times2002$ in which $9$ is not a divisor. Therefore the highest power of $3$ that divides all ten products is $3$ itself.
If $n$ is even then so is $n+2$, and one of them is divisible by $4$ and the other is not, so the product is divisible by $8$. But if $n$ is odd, then so is $n+2$, and $n+1$ is even. Among the cases where $n+1$ is even, we have $n+1=2002$, which is not divisble by $4$, so $4$ is not a common divisor of all ten products. The highest power of $2$ that divides all ten products is therefore $2$ itself.
The next prime number is $5$. Some of these products are not divisible by $5$; for example $2001\times2002\times 2003$ is not. So $5$ is not a common divisor.
$7$ divides $2002$ and $2009$ and nothing between those, so $2003\times2004\times2005$ is not divisible by $7$.
The next prime is $11$. This one divides $2002$ but nothing else until you get up to $2013$, which is not included, so $11$ does not divide $2003\times2004\times 2005$. In a similar way you can rule out all larger primes as common divisors.
Therefore the g.c.d. is $2\times3=6$.
A: Let $n=2000$; the first number is $n^3+3n^2+2n$ and the second number is $(n+1)^3+3(n+1)^2+2(n+1)$, so their difference is
$$
(n+1)^3-n^3+3(n+1)^2-3n^2+2(n+1)-2n=3n^2+9n+6
$$
The third number is $(n+2)^3+3(n+2)^2+2(n+2)$ and the difference with the first number is
$$
(n+2)^3-n^3+3(n+2)^2-3n^2+2(n+2)-2n=6n^2+24n+24
$$
The gcd should be a divisor of
$$
(6n^2+24n+24)-2(3n^2+9n+6)=6n+12=6(n+2)
$$
The same computation from $n=2001$ shows the gcd must be a divisor of both $6\cdot2002$ and $6\cdot2003$, but $2002$ and $2003$ are coprime.
Hence the gcd is a divisor of $6$; now, $n^3+3n^2+2n=n(n+1)(n+2)$, so…
A: The sequence  $\,f_n = n(n\!+\!1)(n\!+\!2)\,$ satisfies $\,\nabla^3 f_n = 6,\,$ where $\,\nabla f_n := f_{n+1}-f_n,\, $ therefore $\,d \mid f_n,f_{n+1},f_{n+2},f_{n+3}\,\Rightarrow\,d\mid \nabla^3 f_n = 6.\,$ Conversely $\ 6\mid f_n = 3!{n+2\choose 3}\ $ for all $\,n.$  
Therefore, more generally $\ k\ge 3\,\Rightarrow\,\gcd(f_n,f_{n+1},\ldots,f_{n+k})= 6 $
