Difference between revisions of "2003 AIME II Problems/Problem 1"

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== Solution ==
 
== Solution ==
Let the three integers be <math>a, b, c</math>.  <math>N = abc = 6(a + b + c)</math> and <math>c = a + b</math>.  Then <math>N = ab(a + b) = 6(a + b + a + b) = 12(a + b)</math>.  Since <math>a</math> and <math>b</math> are positive, <math>ab = 12</math> so <math>\{a, b\}</math> is one of <math>\{1, 12\}, \{2, 6\}, \{3, 4\}</math> so <math>a + b</math> is one of <math>13, 8, 7</math> so <math>N</math> is one of <math>12\cdot 13 = 156, 12\cdot 8 = 96, 12\cdot 7 = 84</math> so the answer is <math>156 + 96 + 84 = 336</math>.
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Let the three integers be <math>a, b, c</math>.  <math>N = abc = 6(a + b + c)</math> and <math>c = a + b</math>.  Then <math>N = ab(a + b) = 6(a + b + a + b) = 12(a + b)</math>.  Since <math>a</math> and <math>b</math> are positive, <math>ab = 12</math> so <math>\{a, b\}</math> is one of <math>\{1, 12\}, \{2, 6\}, \{3, 4\}</math> so <math>a + b</math> is one of <math>13, 8, 7</math> so <math>N</math> is one of <math>12\cdot 13 = 156, 12\cdot 8 = 96, 12\cdot 7 = 84</math> so the answer is <math>156 + 96 + 84 = \boxed{336}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 00:45, 27 February 2011

Problem

The product $N$ of three positive integers is $6$ times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of $N$.

Solution

Let the three integers be $a, b, c$. $N = abc = 6(a + b + c)$ and $c = a + b$. Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$. Since $a$ and $b$ are positive, $ab = 12$ so $\{a, b\}$ is one of $\{1, 12\}, \{2, 6\}, \{3, 4\}$ so $a + b$ is one of $13, 8, 7$ so $N$ is one of $12\cdot 13 = 156, 12\cdot 8 = 96, 12\cdot 7 = 84$ so the answer is $156 + 96 + 84 = \boxed{336}$.

See also

2003 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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