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>. | + | 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 of three positive integers is times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of .
Solution
Let the three integers be . and . Then . Since and are positive, so is one of so is one of so is one of so the answer is .
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
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 |