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 (Basically, I changed the wording to the actual wording of the contest, minus the figure that was provided. Also, this question had no multiple choice answers, so I added those in.)
Line 26: Line 26:
  
 
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>.
 
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>.
 +
{{MAA Notice}}

Revision as of 12:50, 5 July 2013

Problem

Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to him in the circle. Then each person computes and announces the average of the numbers of his two neighbors. The average announced by each person was (in order around the circle) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 (NOT the original number the person picked). The number picked by the person who announced the average 6 was

(A) 1 (B) 5 (C) 6 (D) 10 (E) not uniquely determined from the given information


Solution

Number the people $1$ to $10$ in order in which they announced the numbers. Let $a_i$ be the number chosen by person $i$.

For each $i$, the number $i$ is the average of $a_{i-1}$ and $a_{i+1}$ (indices taken modulo $10$). Or equivalently, the number $2i$ is the sum of $a_{i-1}$ and $a_{i+1}$.

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 $a_6$, we are interested in these equations:

\begin{align} 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}

Summing all five of them, we get $2(a_2 + a_4 + a_6 + a_8 + a_{10}) = 50$, hence $a_2 + a_4 + a_6 + a_8 + a_{10} = 25$.

If we now take the sum of all five variables and subtract equations $(1)$ and $(4)$, we see that $a_6 = 25 - 6 - 18 = \boxed{1}$. 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=56400"