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

Latest revision as of 23:17, 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.) $[asy] unitsize(2 cm); for(int i = 1; i <= 10; ++i) { label("``" + (string) i + "''", dir(90 - 360/10*(i - 1))); } [/asy]$ 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

@@ Line 1: / Line 1: @@
 ==Problem==
-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.
+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>
+unitsize(2 cm);
-What was the number chosen by the girl who announced the number 6?
+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 <math>6</math> was
-{{incomplete|answers}}
+<math>\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}</math>
-==Solution==
+==Solution 1 (Ten Variables)==
-{{solution}}
+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>
-==See also==
+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
+<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) \\
+a_{10} + a_2 & = 2. &&(5)
+\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>
+~Misof (Solution)
+~MRENTHUSIASM (Revision)
+==Solution 2 (One Variable)==
+For <math>i\in\{1,2,3,\ldots,10\},</math> suppose Person <math>i</math> announces the number <math>i.</math>
+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]
+\textbf{People} & \textbf{Average of \#s Picked} & \textbf{Sum of \#s Picked} & \multicolumn{1}{c}{\textbf{Conclusion}} \\ [0.5ex]
+\hline
+& & & \\ [-2ex]
+\text{ and }8 & 7 & 14 & \text{Person 8 picks } 14-x. \\
+\text{ and }10 & 9 & 18 & \text{Person 10 picks } 4+x. \\
+\text{ and }2 & 1 & 2 & \text{Person 2 picks } -2-x \\
+\text{ and }4 & 3 & 6 & \text{Person 4 picks } 8+x \\
+\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>
+~MRENTHUSIASM
+==Video Solution by SpreadTheMathLove==
+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}}

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

Page

Toolbox

Search