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Difference between revisions of "1985 AIME Problems/Problem 3"

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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>.
 
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>.
+
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>.
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 +
=Video Solution by SpreadTheMathLove=
 +
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}}
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* [[AIME Problems and Solutions]]
 
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* [[American Invitational Mathematics Examination]]
 
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* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 17:58, 10 March 2024

Problem

Find $c$ if $a$, $b$, and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$, where $i^2 = -1$.

Solution

Expanding out both sides of the given equation we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$. Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$. Since $a, b$ are integers, this means $b$ is a divisor of 107, which is a prime number. Thus either $b = 1$ or $b = 107$. If $b = 107$, $3a^2 - 107^2 = 1$ so $3a^2 = 107^2 + 1$, but $107^2 + 1$ is not divisible by 3, a contradiction. Thus we must have $b = 1$, $3a^2 = 108$ so $a^2 = 36$ and $a = 6$ (since we know $a$ is positive). Thus $c = 6^3 - 3\cdot 6 = \boxed{198}$.

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
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