Difference between revisions of "1985 AIME Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | + | Find <math>c</math> if <math>a</math>, <math>b</math>, and <math>c</math> are [[positive integer]]s which satisfy <math>c=(a + bi)^3 - 107i</math>, where <math>i^2 = -1</math>. | |
== Solution == | == Solution == | ||
| + | Expanding out both sides of the given [[equation]] we have <math>c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i</math>. Two [[complex number]]s are equal if and only if their [[real part]]s and [[imaginary part]]s are equal, so <math>c = a^3 - 3ab^2</math> and <math>107 = 3a^2b - b^3 = (3a^2 - b^2)b</math>. Since <math>a, b</math> are [[integer]]s, this means <math>b</math> is a [[divisor]] of 107, which is a [[prime number]]. Thus either <math>b = 1</math> or <math>b = 107</math>. If <math>b = 107</math>, <math>3a^2 - 107^2 = 1</math> so <math>3a^2 = 107^2 + 1</math>, but <math>107^2 + 1</math> is not divisible by 3, a contradiction. Thus we must have <math>b = 1</math>, <math>3a^2 = 108</math> so <math>a^2 = 36</math> and <math>a = 6</math> (since we know <math>a</math> is positive). Thus <math>c = 6^3 - 3\cdot 6 = 198</math>. | ||
== See also == | == See also == | ||
| + | * [[1985 AIME Problems/Problem 2 | Previous problem]] | ||
| + | * [[1985 AIME Problems/Problem 4 | Next problem]] | ||
* [[1985 AIME Problems]] | * [[1985 AIME Problems]] | ||
| + | |||
| + | [[Category:Intermediate Complex Numbers Problems]] | ||
Revision as of 12:49, 18 November 2006
Problem
Find 
if 
, 
, and are positive integers which satisfy 
, where 
.
Solution
Expanding out both sides of the given equation we have 
. Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so 
and 
. Since 
are integers, this means is a divisor of 107, which is a prime number. Thus either 
or 
. If , 
so 
, but 
is not divisible by 3, a contradiction. Thus we must have , 
so 
and 
(since we know is positive). Thus 
.
