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

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The number picked by the person who announced the average <math>6</math> was
 
The number picked by the person who announced the average <math>6</math> was
  
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<math>\textbf{(A) }1 \qquad  
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<math>\textbf{(A) } 1 \qquad  
 
\textbf{(B) } 5 \qquad  
 
\textbf{(B) } 5 \qquad  
 
\textbf{(C) } 6 \qquad  
 
\textbf{(C) } 6 \qquad  
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\textbf{(E) }\text{not uniquely determined from the given information}</math>
 
\textbf{(E) }\text{not uniquely determined from the given information}</math>
  
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==Solution 1 (One Variable)==
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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> announces the number <math>i.</math>
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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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Let <math>x</math> be the number picked by Person <math>6.</math> It follows that
+
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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<cmath>\begin{alignat*}{8}
 
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\text{The sum of the numbers picked by Person 6 and Person 8 is 14.}&\implies
 
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\end{alignat*}</cmath>
 
  
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<b>SOLUTION IN PROGRESS</b>
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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
 +
<cmath>\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) \\
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a_{10} + a_2 & = 2. &&(5)
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\end{align*}</cmath>
 +
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>
 +
Subtracting <math>(1)+(4)</math> from <math>(\bigstar),</math> we obtain <math>a_6=\boxed{\textbf{(A) } 1}.</math>
  
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==Solution 2 (Ten Variables)==
+
~Misof (Solution)
  
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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>.
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~MRENTHUSIASM (Revision)
  
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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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==Solution 2 (One Variable)==
−
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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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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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:
 
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<cmath>\begin{align}
 
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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>
 
−
 
 
−
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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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:
 +
<cmath>\begin{array}{c|c|c||l}
 +
& & & \\ [-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]
 +
\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 \\
 +
\end{array}</cmath>
 +
We have <math>x=2-x,</math> from which <math>x=\boxed{\textbf{(A) } 1}.</math>
  
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<math>\fbox{A}</math>
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~MRENTHUSIASM
  
 
== See also ==
 
== See also ==

Revision as of 09:18, 26 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 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

See also

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