Difference between revisions of "2003 AIME II Problems/Problem 1"
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== Problem == | == Problem == | ||
| − | The product <math>N</math> of three positive | + | The [[product]] <math>N</math> of three [[positive integer]]s is <math>6</math> times their [[sum]], and one of the [[integer]]s is the sum of the other two. Find the sum of all possible values of <math>N</math>. |
== 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>. |
== See also == | == See also == | ||
| − | + | {{AIME box|year=2003|before=First Question|num-a=2|n=II}} | |
| − | + | [[Category:Intermediate Algebra Problems]] | |
| + | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 22:00, 25 February 2007
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 | ||
