There is more than one integer greater than $1$ which, when divided by any integer $k$ such that $2 \le k \le 11$, has a remainder of $1$. 
What is the difference between the two smallest such integers?
$\textbf{(A) }2310\qquad \textbf{(B) }2311\qquad \textbf{(C) }27,720\qquad \textbf{(D) }27,721\qquad  \textbf{(E) }\text{none of these}$
Let this integer be $n$. We have
$n\equiv 1 \mod 2$, $n \equiv 1 \mod 3$, $n \equiv 1 \mod 4$ $\ldots$ $n \equiv 1 \mod 11$.
Recall that if 
\[a \equiv b \mod c\]
and \[a \equiv b \mod d\]
then \[a \equiv b \mod (lcm (c,d))\]
We see that since
$n\equiv 1 \mod 2$, $n \equiv 1 \mod 3$, $n \equiv 1 \mod 4$ $\ldots$ $n \equiv 1 \mod 11$.
We have 
\[n \equiv 1 \mod (lcm (2, 3, \ldots , 10, 11))\]
From $2$ to $11$, $8$ contains the largest power of $2$, $9$ contains the largest power of $3$, and $10$ contains the largest power of $5$. Thus, our lcm is equal to
\[2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11\]
Since $n > 1$, our $2$ smallest values of $n$ are 
\[n = 1 + (2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11)\] 
and
\[n = 1 + 2(2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11)\]
The difference between these values is simply the value of \[(2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11) = 40 \cdot 99 \cdot 7\]
\[= 3960 \cdot 7\]
\[= 27720, \boxed{C}\]
~JustinLee2017