2003 AMC 10A Problem 15

Question

What is the probability that an integer in the set

\{1, 2, 3, \ldots, 100\}

is divisible by

2

and not divisible by

3?

\dfrac{1}{6}

\dfrac{33}{100}

\dfrac{17}{50}

\dfrac{1}{2}

\dfrac{18}{25}

Count the multiples of

2,

then remove those also divisible by

3

There are

50

even numbers and

16

multiples of

6

Of the

100

integers,

50

are divisible by

2.

Among those, the ones also divisible by

3

are the multiples of

6,

of which there are

16.

So

50 - 16 = 34

qualify, giving probability

\dfrac{34}{100} = \dfrac{17}{50}.

Thus, the correct answer is C.

Problem 15 in other years