Difference between revisions of "2009 AIME I Problems/Problem 2"
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== Solution == | == Solution == | ||
+ | 1st Solution | ||
Let <math>z = a + 164i</math>. | Let <math>z = a + 164i</math>. | ||
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and this equation shows that <math>n = \boxed{697}.</math> | and this equation shows that <math>n = \boxed{697}.</math> | ||
+ | |||
+ | |||
+ | 2nd solution | ||
+ | |||
+ | <cmath>\frac {z}{z+n}=4i</cmath> | ||
+ | |||
+ | <cmath>1-\frac {n}{z+n}=4i</cmath> | ||
+ | |||
+ | <cmath>1-4i=\frac {n}{z+n}</cmath> | ||
+ | |||
+ | <cmath>\frac {1}{1-4i}=\frac {z+n}{n}</cmath> | ||
+ | |||
+ | <cmath>\frac {1+4i}{17}=\frac {z}{n}-1</cmath> | ||
+ | |||
+ | Since their imaginery part has to be equal, | ||
+ | |||
+ | <cmath>\frac {4i}{17}=\frac {164i}{n}</cmath> | ||
+ | |||
+ | <cmath>n=\frac {(164)(17)}{4}=697</cmath> | ||
+ | |||
+ | <cmath>n = \boxed{697}.</cmath> | ||
== See also == | == See also == |
Revision as of 22:32, 19 March 2009
Problem
There is a complex number with imaginary part and a positive integer such that
Find .
Solution
1st Solution
Let .
Then and
From this, we conclude that and
We now have an equation for :
and this equation shows that
2nd solution
Since their imaginery part has to be equal,
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |