Difference between revisions of "1990 AHSME Problems/Problem 26"
(New page: ==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 ann...) |
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What was the number chosen by the girl who announced the number 6? | What was the number chosen by the girl who announced the number 6? | ||
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==Solution== | ==Solution== | ||
| − | {{ | + | |
| + | Number the girls <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 girl <math>i</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>). | ||
| + | Or equivalently, the number <math>2i</math> is the sum of <math>a_{i-1}</math> and <math>a_{i+1}</math>. | ||
| + | |||
| + | We can split these ten equations into two independent sets of five - one for the even-numbered girls, one for the odd-numbered ones. As we only need <math>a_6</math>, we are interested in these equations: | ||
| + | |||
| + | <cmath>\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}</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>. | ||
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| + | 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>. | ||
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==See also== | ==See also== | ||
Revision as of 22:08, 29 January 2009
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.
What was the number chosen by the girl who announced the number 6?
Solution
Number the girls 
to 
in order in which they announced the numbers. Let 
be the number chosen by girl 
.
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 girls, 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 
.
