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

(→‎See also)
m (Basically, I changed the wording to the actual wording of the contest, minus the figure that was provided. Also, this question had no multiple choice answers, so I added those in.)
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==Problem==
 
==Problem==
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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.
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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
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(A) 1 (B) 5 (C) 6 (D) 10 (E) not uniquely determined from the given information
  
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What was the number chosen by the girl who announced the number 6?
 
  
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{{incomplete|question}}
 
  
 
==Solution==
 
==Solution==
  
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Number the girls <math>1</math> to <math>10</math> in order in which they announced the numbers. Let <math>a_i</math> be the number chosen by girl <math>i</math>.
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Number the people <math>1</math> to <math>10</math> in order in which they announced the numbers. Let <math>a_i</math> be the number chosen by person <math>i</math>.
  
 
For each <math>i</math>, the number <math>i</math> is the average of <math>a_{i-1}</math> and <math>a_{i+1}</math> (indices taken modulo <math>10</math>).
 
For each <math>i</math>, the number <math>i</math> is the average of <math>a_{i-1}</math> and <math>a_{i+1}</math> (indices taken modulo <math>10</math>).
 
Or equivalently, the number <math>2i</math> is the sum of <math>a_{i-1}</math> and <math>a_{i+1}</math>.
 
Or equivalently, the number <math>2i</math> is the sum of <math>a_{i-1}</math> and <math>a_{i+1}</math>.
  
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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 <math>a_6</math>, we are interested in these equations:
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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 <math>a_6</math>, we are interested in these equations:
  
 
<cmath>\begin{align}
 
<cmath>\begin{align}
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If we now take the sum of all five variables and subtract equations <math>(1)</math> and <math>(4)</math>, we see that <math>a_6 = 25 - 6 - 18 = \boxed{1}</math>.
 
If we now take the sum of all five variables and subtract equations <math>(1)</math> and <math>(4)</math>, we see that <math>a_6 = 25 - 6 - 18 = \boxed{1}</math>.
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==See also==
 
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[http://www.online-poker-spielen.biz/ poker spielen]
 

Revision as of 22:44, 10 January 2012

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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