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

(Cleaned up Sol 1 a bit so it is even more professional.)
(→‎Video Solution by SpreadTheMathLove)
 
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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>
 
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>
  
−
Taking the indices modulo <math>10,</math> we are given that <math>a_i=\frac{a_{i-1}+a_{i+1}}{2},</math> from which <math>2a_i=a_{i-1}+a_{i+1}.</math>
+
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>
  
 
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
 
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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a_{10} + a_2 & = 2. &&(5)
 
a_{10} + a_2 & = 2. &&(5)
 
\end{align*}</cmath>
 
\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>
+
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>
 
Subtracting <math>(1)+(4)</math> from <math>(\bigstar),</math> we obtain <math>a_6=\boxed{\textbf{(A) } 1}.</math>
  
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~MRENTHUSIASM
 
~MRENTHUSIASM
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 +
==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=cqtr_OgZ3Xg
  
 
== See also ==
 
== See also ==

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
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All AHSME Problems and Solutions

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