2003 AIME II Problem 1

Question

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.

Answer

Let the integers be

a,

b,

and

c = a + b,

and substitute into

abc = 6(a + b + c)

The equation reduces to

ab = 12;

each way of factoring

12

gives one value of

N = 12c

Let the integers be

a,

b,

and

c = a + b.

Then

\begin{aligned} N &= abc = 6(a + b + c) \\ &= 6 \cdot 2c = 12c, \end{aligned}

and cancelling

c

from

abc = 12c

leaves

ab = 12.

The factorizations

(a, b) = (1, 12),

(2, 6),

(3, 4)

give

c = 13,

8,

7

and

N = 12c = 156,

96,

84.

The sum of all possible values is

156 + 96 + 84 = 336.

Problem 1 in other years