Difference between revisions of "2009 AIME I Problems/Problem 2"

(Solution)
(Solution)
Line 8: Line 8:
  
 
== Solution ==
 
== Solution ==
 +
1st Solution
  
 
Let <math>z = a + 164i</math>.
 
Let <math>z = a + 164i</math>.
Line 24: Line 25:
  
 
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 $z$ with imaginary part $164$ and a positive integer $n$ such that

\[\frac {z}{z + n} = 4i.\]

Find $n$.

Solution

1st Solution

Let $z = a + 164i$.

Then \[\frac {a + 164i}{a + 164i + n} = 4i\] and \[a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.\]

From this, we conclude that \[a = -656\] and \[164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).\]

We now have an equation for $n$: \[4i \left (-656 + n \right ) = 164i,\]

and this equation shows that $n = \boxed{697}.$


2nd solution

\[\frac {z}{z+n}=4i\]

\[1-\frac {n}{z+n}=4i\]

\[1-4i=\frac {n}{z+n}\]

\[\frac {1}{1-4i}=\frac {z+n}{n}\]

\[\frac {1+4i}{17}=\frac {z}{n}-1\]

Since their imaginery part has to be equal,

\[\frac {4i}{17}=\frac {164i}{n}\]

\[n=\frac {(164)(17)}{4}=697\]

\[n = \boxed{697}.\]

See also

2009 AIME I (ProblemsAnswer KeyResources)
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