Understanding the 2003 AIME Problem

The video tutorial addresses a problem from the 2003 AIME exam, focusing on three positive integers whose product is six times their sum, with one integer being the sum of the other two. The instructor defines variables, sets up equations, and simplifies the problem to find possible integer values. The process involves algebraic manipulation and verification of solutions, leading to the calculation of the sum of all possible values of N, which is 336.

What does AIME stand for?

• American Invitational Mathematics Exam

• Association of International Math Experts

• American Institute of Math Education

• Advanced International Math Examination

What is the relationship between the integers a, b, and c?

• a + b = c

• a * b = c

• a = b + c

• a * c = b

What equation is derived from the given conditions?

• a + b + c = 12

• a * b = 12

• a + b = 12

• a * b * c = 12

If a = 1 and b = 12, what is the value of c?

• 11

• 12

• 13

• 14

What is the product of the integers when a = 2 and b = 6?

• 96

• 84

• 156

• 72

What is the sum of the integers when a = 3 and b = 4?

• 10

• 9

• 8

• 7

Why can't the values of a and b be greater than 12?

• They would be zero

• They would not be integers

• They would not satisfy the equation

• They would be negative

What is the product of a and b when a = 3 and b = 4?

• 96

• 84

• 7

• 12

What is the sum of all possible values of N?

• 96

• 84

• 156

• 336

What is the final answer to the problem?

• 156

• 96

• 84

• 336