Secret Number Problem Ten students are seated around a (circular) table. Each student selects his or her own secret number and tells the person on his or her right side the number and the person his or her left side the number (without disclosing it to anyone else). Each student, upon hearing two numbers, then calculates the average and announces it aloud. In order, going around the table, the announced averages are 1,2,3,4,5,6,7,8,9 and 10. What was the secret number chosen by the person who announced a 6?
 A: Let us denote secret numbers as $x_i$ , where $i$ is announced number ,then we have following system of equations :
$\begin{cases}
 x_1+x_3=4 \\
 x_2+x_4=6 \\
 x_3+x_5=8 \\
 x_4+x_6=10 \\
 x_5+x_7=12 \\
 x_6+x_8=14 \\
 x_7+x_9=16 \\
 x_8+x_{10}=18 \\
 x_9+x_1=20 \\
 x_{10}+x_2=2
\end{cases}$
According to Maple : $x_6=1$ , so requested secret number is $1$ .
A: Extending pedja's analysis:
Let us denote secret numbers as $x_i$ , where $i$ is announced number ,then we have following system of equations :
$\begin{cases}
 x_1+x_3=4 \\
 x_2+x_4=6 \\
 x_3+x_5=8 \\
 x_4+x_6=10 \\
 x_5+x_7=12 \\
 x_6+x_8=14 \\
 x_7+x_9=16 \\
 x_8+x_{10}=18 \\
 x_9+x_1=20 \\
 x_{10}+x_2=2
\end{cases}$
So we have 
$(x_6+x_8)-(x_8+x_{10})+(x_{10}+x_2)-(x_2+x_4)+(x_4+x_6) = \\
 x_6+(x_8-x_8)+(-x_{10}+x_{10})+(x_2-x_2)+(-x_4+x_4)+x_6 = x_6 + x_6$
$14-18+2-6+10 = 2 = 2x_6$
$x_6 = 1$
Maple not required. 
A: Hint:  If we label the chosen numbers $a,b,c,d,e,f,g,h,i$ you only care about the even position ones (why?)  So you get a system $b+d=2\cdot 3=6, d+f=2\cdot 5=10, \ldots$.  Five equations in five unknowns.
