Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem)
(→‎Video Solution by SpreadTheMathLove)
 
(8 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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 (<i>not</i> the original number the person picked.)
+
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>
 
<asy>
 
unitsize(2 cm);
 
unitsize(2 cm);
Line 8: Line 8:
 
}
 
}
 
</asy>
 
</asy>
−
The number picked by the person who announced the average 6 was
+
The number picked by the person who announced the average <math>6</math> was
  
−
<math>\textbf{(A) }1 \qquad  
+
<math>\textbf{(A) } 1 \qquad  
 
\textbf{(B) } 5 \qquad  
 
\textbf{(B) } 5 \qquad  
 
\textbf{(C) } 6 \qquad  
 
\textbf{(C) } 6 \qquad  
Line 16: Line 16:
 
\textbf{(E) }\text{not uniquely determined from the given information}</math>
 
\textbf{(E) }\text{not uniquely determined from the given information}</math>
  
−
==Solution==
+
==Solution 1 (Ten Variables)==
 +
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>
  
−
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>.
+
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>
  
−
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>).
+
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
−
Or equivalently, the number <math>2i</math> is the sum of <math>a_{i-1}</math> and <math>a_{i+1}</math>.
+
<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>
  
−
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:
+
~Misof (Solution)
  
−
<cmath>\begin{align}
+
~MRENTHUSIASM (Revision)
−
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}</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>.
+
==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>
  
−
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>.
+
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]
 +
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}</cmath>
 +
We have <math>x=2-x,</math> from which <math>x=\boxed{\textbf{(A) } 1}.</math>
  
−
<math>\fbox{A}</math>
+
~MRENTHUSIASM
 +
 
 +
==Video Solution by SpreadTheMathLove==
 +
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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_26&oldid=198552"