2010 AMC 12B Problem 16

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}

Factor

abc+ab+a=a(bc+b+1);

each residue mod

3

is equally likely

If

a

is divisible by

3

it works; otherwise require

bc+b+1

to be divisible by

3

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

a

is divisible by

3

(probability

\tfrac13

), the product is divisible by

3.

If

a

is not divisible by

3

(probability

\tfrac23

), we need

bc+b+1

to be divisible by

3.

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.

Problem 16 in other years