If $n$ is any integer, $n+3, n-9, n-4, n+6$, and $n-1$ are also integers. If $n+3, n-9, n-4, n+6$, and $n-1$ are arranged from smallest to largest, the integer in the middle is
(A) $n+3$
(B) $n-9$
(C) $n-4$
(D) $n+6$
(E) $n-1$
Since $n$ can be any integer, let us choose $n=0$. Then the values of the 5 integers are 3 , $-9,-4,6$, and -1 . When we arrange these from smallest to largest, we get $-9,-4,-1$, 3 , and 6 , so the number in the middle is -1 , which is $n-1$.