A data set consists of $6$ (not distinct) positive integers: $1$, $7$, $5$, $2$, $5$, and $X$. The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all possible values of $X$?
$\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40$
First, note that $1+7+5+2+5=20$. There are $3$ possible cases:
Case 1: the mean is $5$.
$X = 5 \cdot 6 - 20 = 10$.
Case 2: the mean is $7$.
$X = 7 \cdot 6 - 20 = 22$.
Case 3: the mean is $X$.
$X= \frac{20+X}{6} \Rightarrow X=4$.
Therefore, the answer is $10+22+4=\boxed{\textbf{(D) }36}$.
~MrThinker