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 | + | 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 the sum of all possible values of <math>N</math> is <math>12 \cdot (13 + 8 + 7) = 12(28) = \boxed{336}</math>. |
== Video Solution by Sal Khan == | == Video Solution by Sal Khan == | ||
Latest revision as of 07:56, 11 July 2023
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 the sum of all possible values of is 
.
Video Solution by Sal Khan
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. 
