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 $N$ is one of $12\cdot 13 = 156, 12\cdot 8 = 96, 12\cdot 7 = 84$ so the answer is $156 + 96 + 84 = \boxed{336}$ .

Video Solution by Sal Khan

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

2003 AIME II (Problems • Answer Key • Resources)
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2003 AIME II Problems/Problem 1

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Problem

Solution

Video Solution by Sal Khan

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