1990 AHSME Problems/Problem 26

Problem

Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to them in the circle. Then each person computes and announces the average of the numbers of their two neighbors. The figure shows the average announced by each person (not the original number the person picked.) $[asy] unitsize(2 cm); for(int i = 1; i <= 10; ++i) { label("``" + (string) i + "''", dir(90 - 360/10*(i - 1))); } [/asy]$ The number picked by the person who announced the average $6$ was

$\textbf{(A) } 1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 10 \qquad \textbf{(E) }\text{not uniquely determined from the given information}$

Solution 1 (Ten Variables)

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

$\fbox{A}$

Solution 2 (One Variable)

For $i\in\{1,2,3,\ldots,10\},$ suppose Person $i$ announces the number $i.$

Let $x$ be the number picked by Person $6.$ We construct the following table: $\begin{array}{c|c|c||l} & & & \\ [-2.5ex] \textbf{People} & \textbf{Average of \#s Picked} & \textbf{Sum of \#s Picked} & \multicolumn{1}{c}{\textbf{Conclusion}} \\ [0.5ex] \hline & & & \\ [-2ex] 6\text{ and }8 & 7 & 14 & \text{Person 8 picks } 14-x. \\ 8\text{ and }10 & 9 & 18 & \text{Person 10 picks } 4+x. \\ 10\text{ and }2 & 1 & 2 & \text{Person 2 picks } -2-x \\ 2\text{ and }4 & 3 & 6 & \text{Person 4 picks } 8+x \\ 4\text{ and }6 & 5 & 10 & \text{Person 6 picks } 2-x \\ \end{array}$ We have $x=2-x,$ from which $x=\boxed{\textbf{(A) } 1}.$

~MRENTHUSIASM

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1990 AHSME Problems/Problem 26

Contents

Problem

Solution 1 (Ten Variables)

Solution 2 (One Variable)

See also