What's up with A?

Let $S = \{1, 2, 3, 4, 5\}$ .

The sum of 3 distinct numbers from S can at least be $1+2+3 = 6$ and at most be $3+4+5 = 12$ , leaving the possible range for average to be $[2, 3, 4]$ This process is similar with the sum of 2 distinct numbers from S. (We are only considering integers because all elements of S are integers.) We now know that we can only choose B and C from ${2, 3, 4}$ .

Considering statement C in addition, this gives the configuration of {A, B, C} several possibilities: $\{1, 3, 2\}, \{2, 4, 3\}, \{4, 2, 3\}, \{5, 3, 4\}$

Considering statement B, those configurations could be extended to {A, B, C, D} as follows, respectively: $\{1, 3, 2, 4\}, \{2, 4, 3, 5\}, \{4, 2, 3, 1\}, \{5, 3, 4, 2\}$

Considering statement D, $\{1, 3, 2, 4\}, \{2, 4, 3, 5\}$ remain valid. Those can be extended to configurations of {A, B, C, D, E}: $\{1, 3, 2, 4, 5\}, \{2, 4, 3, 5, 1\}$

Considering statement E, $\{2, 4, 3, 5, 1\}$ is the correct configuration of $\{A, B, C, D, E\}$ . Therefore, statement A corresponds to $\boxed{2}$ .

Since $B = \frac {C+D}2$ and $D = B+1$ , $C = B-1$ . This leaves the possibilities $(B,C,D) = (2,1,3), (3,2,4), (4,3,5)$ .

Because A and E are consecutive, $(B,C,D) = (3,2,4)$ isn't possible.

For the two remaining options, $C = \frac {A+B+C}3 \Leftrightarrow A = 2C-B$ , so A either equals 0 or 2, but only $A=2$ satisfies E's condition.

To sum up, $(A,B,C,D,E) = (2,4,3,5,1)$ and so the answer is 2 .