Difference between revisions of "1990 AHSME Problems/Problem 26"
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==Problem== | ==Problem== | ||
− | Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to | + | 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); |
+ | 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 | ||
+ | <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== |
Revision as of 05:33, 9 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 ([i]not[/i] the original number the person picked.)
The number picked by the person who announced the average 6 was
Solution
Number the people to
in order in which they announced the numbers. Let
be the number chosen by person
.
For each , the number
is the average of
and
(indices taken modulo
).
Or equivalently, the number
is the sum of
and
.
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 , we are interested in these equations:
Summing all five of them, we get , hence
.
If we now take the sum of all five variables and subtract equations and
, we see that
.
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.