smc

Let this integer be $n$. We have $n\equiv 1\mathrm{mod}2$, $n\equiv 1\mathrm{mod}3$, $n\equiv 1\mathrm{mod}4$ $\dots$ $n\equiv 1\mathrm{mod}11$. Recall that if $$a\equiv b\mathrm{mod}c$$ and $$a\equiv b\mathrm{mod}d$$ then $$a\equiv b\mathrm{mod}(lcm(c,d))$$ We see that since $n\equiv 1\mathrm{mod}2$, $n\equiv 1\mathrm{mod}3$, $n\equiv 1\mathrm{mod}4$ $\dots$ $n\equiv 1\mathrm{mod}11$. We have $$n\equiv 1\mathrm{mod}(lcm(2,3,\dots ,10,11))$$ 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 $$(2^3\cdot 3^2\cdot 5\cdot 7\cdot 11)=40\cdot 99\cdot 7$$ $$=3960\cdot 7$$ $$=27720,\boxed{27,720}$$