Problem:
Positive integers a,b, and c are randomly and independently selected with replacement from the set {1,2,3,…,2010}. What is the probability that abc+ ab+a is divisible by 3 ?
Answer Choices:
A. 31​
B. 8129​
C. 8131​
D. 2711​
E. 2713​
Solution:
Let N=abc+ab+a=a(bc+b+1). If a is divisible by 3 , then N is divisible by 3 . Note that 2010 is divisible by 3 , so the probability that a is divisible by 3 is 31​.
If a is not divisible by 3 then N is divisible by 3 if bc+b+1 is divisible by 3 . Define b0​ and b1​ so that b=3b0​+b1​ is an integer and b1​ is equal to 0,1 , or 2 . Note that each possible value of b1​ is equally likely. Similarly define c0​ and c1​. Then
bc+b+1=(3b0​+b1​)(3c0​+c1​)+3b0​+b1​+1=3(3b0​c0​+c0​b1​+c1​b0​+b0​)+b1​c1​+b1​+1​
Hence bc+b+1 is divisible by 3 if and only if b1​=1 and c1​=1, or b1​=2 and c1​=0. The probability of this occurrence is 31​⋅31​+31​⋅31​=92​.
Therefore the requested probability is 31​+32​⋅92​=2713​​.
The problems on this page are the property of the MAA's American Mathematics Competitions