What is the chosen number of the person who announced an average of 6?

Question

Ten people stand in a circle. Each chooses a number and tells the person to their left and right. Each person averages the two numbers they heard and they go in a circle announcing the average of their numbers which are recorded in this “circle.” What number did the person who had an average of 6 choose?

Although this problem can be solved with a system, is there any logical way to find the answer without a calculator and minimal work? I’ve been trying to figure the answer out and apparently there is an ‘elegant’ solution, but I don’t know what it is.

Thanks.!

Edit: here’s the work if I solved the problem as a system variables per person

Answer 1

This question was also asked here, where many solutions are given. Here is mine.

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Let $g_n$ be the number the $n$th person picked, so that $(g_{n-1}+g_{n+1})/2=n$ for $1\le n\le 10$ (where indices wrap around the circle, so $g_0=g_{10})$. Then $$ \begin{align} g_6 &= (g_6-g_4)/2 + (g_6+g_4)/2\\ &= (g_6+(g_8-g_8)+(-g_{10}+g_{10})+(g_2-g_2)-g_4)/2 + (g_6+g_4)/2\\ &= (g_6+g_8)/2 -(g_8+g_{10})/2+(g_{10}+g_2)/2-(g_2+g_4)/2+(g_6+g_4)/2\\ &= 7-9+1-3+5=\boxed{1} \end{align} $$

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The insight that leads to this proof is that in the circular arrangement $(g_2,g_4,g_6,g_8,g_{10})$, you know what the sum of any two adjacent numbers are. You can tile this circle with "dominos" in two ways; one where every number is covered once except for $g_6$ which is covered twice, and one where every number is covered once except for $g_6$ which is uncovered. Subtracting these two tilings gives you the value of $g_6$.

Answer 2

First, note that the odd and even slots are uncoupled; the even averages only depend on the numbers chosen by the people announcing odd averages and vice versa. Thus, we only need to consider the announced averages $1,3,5,7,9$ to find the number chosen by the person announcing $6$.

Second, we can form this inverse to get an idea how to invert the averaging operation: To find the number chosen between two averages, add them, subtract their neighbours and add their neighbour. This yields $-3,9,1,13,5$, which can readily be checked to yield the above averages, so the number chosen by the person who announced $6$ is $1$.

Answer 3

Here's a nice fact: we can totally ignore considering what any person who announced an odd number said. Why? Because no odd number person is 2 places away from an even number person, so they won't affect each other in the computation of averages anywhere.

Denote the person who said 2 as A, 4, as B, 6 as C, 8 as D, and 10 as E

So,

\begin{align} A+E = 2\\ B+A = 6\\ C+B = 10\\ D+C = 14\\ E+D = 18 \end{align}

After some algebra, you get $A = -3, B= 9, C= 1, D= 13, E= 5$. So the answer is $\color{red}{1}$.

Answer 4

First, the sum of the numbers of the people who announced even numbers is $1+3+5+7+9=25$ The naive approach is to let the person who announced $2$ have the number $a$, the person who announced $4$ have the number $b$ and so on. We want $c$ and get $$a+b=6\\b+c=10\\c+d=14\\d+e=18\\a+e=2\\e=2-a\\d-a=16\\b+d=22\\b-c=8\\c=1\\b=9\\d=13\\a=-3\\e=5$$ where we could have stopped after the $c=1$ line but I did the rest to check. It is only five lines after writing the given equations, each of which is only a couple additions or subtractions.