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2003 AIME II Problems/Problem 1

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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 = 336.

See also

2003 AIME II (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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