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

Revision as of 21:42, 20 March 2009

There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that

$\frac {z}{z + n} = 4i.$

Find $n$ .

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,

$\frac {4i}{17}=\frac {164i}{n}$

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

$n = \boxed{697}.$

Revision as of 20:07, 20 March 2009 (view source) Motoe123 (talk \| contribs) m (→‎Solution) ← Older edit		Revision as of 21:42, 20 March 2009 (view source) Moplam (talk \| contribs) (→‎See also) Newer edit →
Line 50:		Line 50:
	{{AIME box\|year=2009\|n=I\|num-b=1\|num-a=3}}		{{AIME box\|year=2009\|n=I\|num-b=1\|num-a=3}}

−	[[Category:Complex ~~Numbers~~]]	+	[[Category:Complex Number]]

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