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(3b0c0+c0b1+c1b0+b0)+b1c1+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