AMC 10 Step by Step
Log inSign up
Printable

2002 AMC 10A Problem 9

Include the solutionBack to the problem
2002 AMC 10A Problem 9AMC 10 Step by Step
Name:                             Date: September 13, 2026
Systems of Equations · difficulty 2 of 5 · about 2 min · position P1-10

There are 3 numbers A, B, and C, such that 1001C−2002A=40041001C - 2002A = 4004 , and 1001B+3003A=50051001B + 3003A = 5005 . What is the average of A, B, and C?

  1. A)

    11

  2. B)

    33

  3. C)

    66

  4. D)

    99

  5. E)

    Not uniquely determined\text{Not uniquely determined}

Working