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Difference between revisions of "1990 AHSME Problems/Problem 26"

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==Problem==
 
==Problem==
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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 ([i]not[/i] the original number the person picked.)
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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 (<i>not</i> the original number the person picked.)
 
<asy>
 
<asy>
 
unitsize(2 cm);
 
unitsize(2 cm);

Revision as of 04:34, 9 September 2021

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 + "&#039;&#039;", 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

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

$\fbox{A}$

See also

1990 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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