2007 AMC 8 Problem 24

Question

A bag contains four pieces of paper, each labeled with one of the digits

1,

2,

3

or

4,

with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of

3?

\dfrac{1}{4}

\dfrac{1}{3}

\dfrac{1}{2}

\dfrac{2}{3}

\dfrac{3}{4}

Answer: C

Concepts:divisibility basic probability digits

Difficulty rating: 1580

Small hint:

A number is divisible by

3

exactly when its digit sum is divisible by

3.

Big hint:

Instead of ordering the three drawn digits, look at which digit is left out.

Solution:

Recall that a number is divisible by

3

if the sum of its digits is divisible by

3

.

The only triples of distinct digits from

1,2,3,4

whose sum is divisible by

3

are

(1,2,3)

and

(2,3,4)

.

This means the constructed number is a multiple of

3

exactly when either

1

or

4

is left in the bag.

Each of these events has probability

\dfrac{1}{4}

, for a total probability of

2\cdot\dfrac{1}{4}=\dfrac{1}{2}

.

Thus, C is the correct answer.