2003 AMC 10A Problems/Problem 15

Problem

What is the probability that an integer in the set $\{1,2,3,...,100\}$ is divisible by $2$ and not divisible by $3$ ?

$\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{33}{100}\qquad \mathrm{(C) \ } \frac{17}{50}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{18}{25}$

Solution

There are $100$ integers in the set.

Since every 2nd integer is divisible by $2$ , there are $\lfloor\frac{100}{2}\rfloor=50$ integers divisible by $2$ in the set.

To be divisible by both $2$ and $3$ , a number must be divisible by $lcm(2,3)=6$ .

Since every 6th integer is divisible by $6$ , there are $\lfloor\frac{100}{6}\rfloor=16$ integers divisible by both $2$ and $3$ in the set.

So there are $50-16=34$ integers in this set that are divisible by $2$ and not divisible by $3$ .

Therefore, the desired probability is $\frac{34}{100}=\frac{17}{50} \Rightarrow C$

Controversy

Due to the wording of the question, it may be taken as "Find the probability that an integer in said set is divisible by 2 and not 3 EXISTS". One example would be 2, which is not a multiple of 3, thus the probability is 1.

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2003 AMC 10A Problems/Problem 15

Contents

Problem

Solution

Controversy

See Also