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 == | ||
+ | https://www.youtube.com/watch?v=JPQ8cfOsYxo&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=7 | ||
+ | - AMBRIGGS | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 08:56, 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.