3 2 6 2
6
For each way to choose terms of $A$, the average is obtained as follows:
* If just $a_1$ is chosen, the average is $\frac{a_1}{1}=\frac{2}{1} = 2$, which is an integer.
* If just $a_2$ is chosen, the average is $\frac{a_2}{1}=\frac{6}{1} = 6$, which is an integer.
* If just $a_3$ is chosen, the average is $\frac{a_3}{1}=\frac{2}{1} = 2$, which is an integer.
* If $a_1$ and $a_2$ are chosen, the average is $\frac{a_1+a_2}{2}=\frac{2+6}{2} = 4$, which is an integer.
* If $a_1$ and $a_3$ are chosen, the average is $\frac{a_1+a_3}{2}=\frac{2+2}{2} = 2$, which is an integer.
* If $a_2$ and $a_3$ are chosen, the average is $\frac{a_2+a_3}{2}=\frac{6+2}{2} = 4$, which is an integer.
* If $a_1$, $a_2$, and $a_3$ are chosen, the average is $\frac{a_1+a_2+a_3}{3}=\frac{2+6+2}{3} = \frac{10}{3}$, which is not an integer.
Therefore, $6$ ways satisfy the condition.5 5 5 5 5 5
31 Regardless of the choice of one or more terms of $A$, the average equals $5$.
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