Difference between revisions of "1990 AHSME Problems/Problem 26"
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MRENTHUSIASM (talk | contribs) (Cleaned up Sol 1 a bit so it is even more professional.) |
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==Solution 1 (Ten Variables)== | ==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> | ||
− | + | 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> | |
− | + | 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) | |
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==Solution 2 (One Variable)== | ==Solution 2 (One Variable)== |
Revision as of 01:35, 10 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.)
The number picked by the person who announced the average
was
Solution 1 (Ten Variables)
For suppose Person
picks the number
and announces the number
We wish to find
Taking the indices modulo we are given that
from which
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 is
Summing these five equations, we get
, from which
Subtracting
from
we obtain
~Misof (Solution)
~MRENTHUSIASM (Revision)
Solution 2 (One Variable)
For suppose Person
announces the number
Let be the number picked by Person
We construct the following table:
We have
from which
~MRENTHUSIASM
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.