1990 AHSME Problems/Problem 26

Problem

Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to him in the circle. Then each person computes and announces the average of the numbers of his two neighbors. The average announced by each person was (in order around the circle) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 (NOT the original number the person picked). The number picked by the person who announced the average 6 was

(A) 1 (B) 5 (C) 6 (D) 10 (E) not uniquely determined from the given information

Solution

Number the people $1$ to $10$ in order in which they announced the numbers. Let $a_i$ be the number chosen by person $i$ .

For each $i$ , the number $i$ is the average of $a_{i-1}$ and $a_{i+1}$ (indices taken modulo $10$ ). Or equivalently, the number $2i$ is the sum of $a_{i-1}$ and $a_{i+1}$ .

We can split these ten equations into two independent sets of five - one for the even-numbered peoples, one for the odd-numbered ones. As we only need $a_6$ , we are interested in these equations:

$\begin{align} a_2 + a_4 & = 6 \\ a_4 + a_6 & = 10 \\ a_6 + a_8 & = 14 \\ a_8 + a_{10} & = 18 \\ a_{10} + a_2 & = 2 \end{align}$

Summing all five of them, we get $2(a_2 + a_4 + a_6 + a_8 + a_{10}) = 50$ , hence $a_2 + a_4 + a_6 + a_8 + a_{10} = 25$ .

If we now take the sum of all five variables and subtract equations $(1)$ and $(4)$ , we see that $a_6 = 25 - 6 - 18 = \boxed{1}$ .

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1990 AHSME Problems/Problem 26

Problem

Solution