1990 AHSME Problems/Problem 26

Problem

Each of ten girls around a circle chooses a number and tells it to the neighbor on each side. Thus each person gives out one number and receives two numbers. Each girl then announced the average of the two numbers she received. Remarkably, the announced numbers, in order around the circle, were 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

What was the number chosen by the girl who announced the number 6?

Template:Incomplete

Solution

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

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 girls, 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 $a_6 = 25 - 6 - 18 = \boxed{1}$ .

Page

Toolbox

Search

1990 AHSME Problems/Problem 26

Problem

Solution

See also