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

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(→‎Video Solution by SpreadTheMathLove)
 
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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 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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<asy>
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unitsize(2 cm);
  
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What was the number chosen by the girl who announced the number 6?
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for(int i = 1; i <= 10; ++i) {
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  label("``" + (string) i + "&#039;&#039;", dir(90 - 360/10*(i - 1)));
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}
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</asy>
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The number picked by the person who announced the average <math>6</math> was
  
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{{incomplete|question}}
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<math>\textbf{(A) } 1 \qquad
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\textbf{(B) } 5 \qquad
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\textbf{(C) } 6 \qquad
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\textbf{(D) } 10 \qquad
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\textbf{(E) }\text{not uniquely determined from the given information}</math>
  
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==Solution==
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==Solution 1 (Ten Variables)==
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For <math>i\in\{1,2,3,\ldots,10\},</math> suppose Person <math>i</math> picks the number <math>a_i</math> and announces the number <math>i.</math> We wish to find <math>a_6.</math>
  
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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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Taking the indices modulo <math>10,</math> we are given that <math>\frac{a_{i-1}+a_{i+1}}{2}=i,</math> from which <math>a_{i-1}+a_{i+1}=2i.</math>
  
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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>).
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We have ten equations: five with odd-numbered indices and five with even-numbered indices. Note that these two sets of equations are independent. The set that involves <math>a_6</math> is
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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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<cmath>\begin{align*}
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a_2 + a_4 & = 6, &&(1) \\
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a_4 + a_6 & = 10, &&(2) \\
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a_6 + a_8 & = 14, &&(3) \\
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a_8 + a_{10} & = 18, &&(4) \\
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a_{10} + a_2 & = 2. &&(5)
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\end{align*}</cmath>
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Summing these five equations, we get <math>2(a_2 + a_4 + a_6 + a_8 + a_{10}) = 50,</math> from which <cmath>a_2 + a_4 + a_6 + a_8 + a_{10} = 25. \hspace{15mm} (\bigstar)</cmath>
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Subtracting <math>(1)+(4)</math> from <math>(\bigstar),</math> we obtain <math>a_6=\boxed{\textbf{(A) } 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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~Misof (Solution)
  
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<cmath>\begin{align}
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~MRENTHUSIASM (Revision)
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a_2 + a_4 & = 6 \\
 
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a_4 + a_6 & = 10 \\
 
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a_6 + a_8 & = 14 \\
 
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a_8 + a_{10} & = 18 \\
 
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a_{10} + a_2 & = 2
 
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\end{align}</cmath>
 
  
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Summing all five of them, we get <math> 2(a_2 + a_4 + a_6 + a_8 + a_{10}) = 50</math>, hence <math>a_2 + a_4 + a_6 + a_8 + a_{10} = 25</math>.
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==Solution 2 (One Variable)==
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For <math>i\in\{1,2,3,\ldots,10\},</math> suppose Person <math>i</math> announces the number <math>i.</math>
  
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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>.
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Let <math>x</math> be the number picked by Person <math>6.</math> We construct the following table:
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<cmath>\begin{array}{c|c|c||l}
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& & & \\ [-2.5ex]
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\textbf{People} & \textbf{Average of \#s Picked} & \textbf{Sum of \#s Picked} & \multicolumn{1}{c}{\textbf{Conclusion}} \\ [0.5ex]
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\hline
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& & & \\ [-2ex]
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6\text{ and }8 & 7 & 14 & \text{Person 8 picks } 14-x. \\
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8\text{ and }10 & 9 & 18 & \text{Person 10 picks } 4+x. \\
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10\text{ and }2 & 1 & 2 & \text{Person 2 picks } -2-x \\
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2\text{ and }4 & 3 & 6 & \text{Person 4 picks } 8+x \\
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4\text{ and }6 & 5 & 10 & \text{Person 6 picks } 2-x \\
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\end{array}</cmath>
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We have <math>x=2-x,</math> from which <math>x=\boxed{\textbf{(A) } 1}.</math>
  
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~MRENTHUSIASM
  
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==See also==
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==Video Solution by SpreadTheMathLove==
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[http://www.online-poker-spielen.biz/ poker spielen]
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https://www.youtube.com/watch?v=cqtr_OgZ3Xg
 +
 
 +
== See also ==
 +
{{AHSME box|year=1990|num-b=25|num-a=27}} 
 +
 
 +
[[Category: Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:17, 23 September 2023

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 1 (Ten Variables)

For $i\in\{1,2,3,\ldots,10\},$ suppose Person $i$ picks the number $a_i$ and announces the number $i.$ We wish to find $a_6.$

Taking the indices modulo $10,$ we are given that $\frac{a_{i-1}+a_{i+1}}{2}=i,$ from which $a_{i-1}+a_{i+1}=2i.$

We have ten equations: five with odd-numbered indices and five with even-numbered indices. Note that these two sets of equations are independent. The set that involves $a_6$ is \begin{align*} a_2 + a_4 & = 6, &&(1) \\ a_4 + a_6 & = 10, &&(2) \\ a_6 + a_8 & = 14, &&(3) \\ a_8 + a_{10} & = 18, &&(4) \\ a_{10} + a_2 & = 2. &&(5) \end{align*} Summing these five equations, we get $2(a_2 + a_4 + a_6 + a_8 + a_{10}) = 50,$ from which \[a_2 + a_4 + a_6 + a_8 + a_{10} = 25. \hspace{15mm} (\bigstar)\] Subtracting $(1)+(4)$ from $(\bigstar),$ we obtain $a_6=\boxed{\textbf{(A) } 1}.$

~Misof (Solution)

~MRENTHUSIASM (Revision)

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=cqtr_OgZ3Xg

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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