Difference between revisions of "1985 AIME Problems/Problem 3"
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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 = \boxed{198}</math>. | 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 = \boxed{198}</math>. | ||
| − | + | =Video Solution by SpreadTheMathLove= | |
https://www.youtube.com/watch?v=mw2A1Fa7APM | https://www.youtube.com/watch?v=mw2A1Fa7APM | ||
| + | |||
== See also == | == See also == | ||
{{AIME box|year=1985|num-b=2|num-a=4}} | {{AIME box|year=1985|num-b=2|num-a=4}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Latest revision as of 17:58, 10 March 2024
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 
.
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=mw2A1Fa7APM
See also
| 1985 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
