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 = \boxed{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>. | ||
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| + | == Video Solution == | ||
| + | https://www.youtube.com/watch?v=JPQ8cfOsYxo&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=7 | ||
| + | - AMBRIGGS | ||
== See also == | == See also == | ||
Revision as of 17:28, 30 July 2022
Contents
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 
.
Video Solution
https://www.youtube.com/watch?v=JPQ8cfOsYxo&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=7 - AMBRIGGS
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
