2003 AIME II Problems/Problem 1

Problem

The product $N$ of three positive integers is $6$ times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of $N$.

Solution

Let the three integers be $a, b, c$. $N = abc = 6(a + b + c)$ and $c = a + b$. Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$. Since $a$ and $b$ are positive, $ab = 12$ so $\{a, b\}$ is one of $\{1, 12\}, \{2, 6\}, \{3, 4\}$ so $a + b$ is one of $13, 8, 7$ so the sum of all possible values of $N$ is $12 \cdot (13 + 8 + 7) = 12(28) = \boxed{336}$.

Video Solution by Sal Khan

https://www.youtube.com/watch?v=JPQ8cfOsYxo&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=7 - AMBRIGGS

See also

2003 AIME II (ProblemsAnswer KeyResources)
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

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