The sum of four consecutive odd integers is 216. Find the four integers.

A) $54,53,55,57$
B) $53,55,57,51$
C) $50,53,55,57$
D) $51,53,55,57$
E) $52,53,55,57$

Let the four consecutive odd integers be $n$, $n+2$, $n+4$, and $n+6$. Then the sum of these integers is $4n+12$. Setting this equal to 216 and solving for $n$, we get $n=51$. Therefore, the four integers are 51, 53, 55, and 57, which corresponds to choice (D).