Difference between revisions of "2009 AIME I Problems/Problem 2"
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<cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath> | <cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath> | ||
− | By comparing coefficients, equating the real terms on | + | By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation, |
we conclude that | we conclude that |
Revision as of 22:08, 23 February 2012
Problem
There is a complex number with imaginary part
and a positive integer
such that
Find .
Solution
1st Solution
Let .
Then
and
By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,
we conclude that
By equating the imaginary terms on each side of the equation,
we conclude that
We now have an equation for :
and this equation shows that
2nd Solution
Since their imaginary 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 |