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 $\frac{34}{100}=\frac{17}{50} \Rightarrow C$

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2003 AMC 10A Problems/Problem 15

Problem

Solution

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