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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?

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  • $\begingroup$ Hint: think about it as a linear system. $\endgroup$ – savick01 Jan 8 '12 at 17:00
  • $\begingroup$ are the numbers real, integer or positive integers? $\endgroup$ – N. S. Jan 8 '12 at 17:01
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    $\begingroup$ Here's how to do most of these problems. Look at cases where there are only three people, then four people, then work your way up :) $\endgroup$ – itdoesntwork Jan 8 '12 at 17:14
  • $\begingroup$ @itdoesntwork: this is impossible with 4 people: the students on opposite corners are getting fed the same numbers! $\endgroup$ – user641 Jan 8 '12 at 18:59
  • $\begingroup$ @SteveD Right you are! I didn't think that one through that much :) $\endgroup$ – itdoesntwork Jan 9 '12 at 0:44
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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 $1$ .

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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.

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  • $\begingroup$ +1 This shows why you only care about the even positions-the odd ones add out. The person who announced 6 has extra information about them. $\endgroup$ – Ross Millikan May 17 '12 at 5:26
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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.

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