Problem:
Suppose A,B, and C are three numbers for which 1001C−2002A=4004 and 1001B+3003A=5005. The average of the three numbers A,B, and C is
Answer Choices:
A. 1
B. 3
C. 6
D. 9
E. not uniquely determined
Solution:
Adding 1001C−2002A=4004 and 1001B+3003A=5005 yields 1001A+ 1001B+1001C=9009. So A+B+C=9, and the average is
3A+B+C​=3.
Answer: B​.
The problems on this page are the property of the MAA's American Mathematics Competitions