Question

Positive integers $a, b,$ and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3, \ldots, 2010\}.$ What is the probability that $abc+ab+a$ is divisible by $3?$

$\dfrac{1}{3}$

$\dfrac{29}{81}$

$\dfrac{31}{81}$

$\dfrac{11}{27}$

$\dfrac{13}{27}$

Solution:

Factor $abc+ab+a=a(bc+b+1).$ Since $2010$ is a multiple of $3,$ each of $a, b, c$ is uniform modulo $3.$

If $3\mid a$ (probability $\tfrac13$ ), the product is divisible by $3.$

If $3\nmid a$ (probability $\tfrac23$ ), we need $3\mid bc+b+1.$ Checking residues, this holds exactly when $(b,c)\equiv(1,1)$ or $(2,0)\pmod3,$ a probability of $\tfrac13\cdot\tfrac13+\tfrac13\cdot\tfrac13=\tfrac29.$

The total probability is $\frac13+\frac23\cdot\frac29=\frac13+\frac{4}{27}=\frac{13}{27}.$

Thus, the correct answer is E.

2010 AMC 12B Problem 16

Solution:

Problem 16 in Other Years