Probability in Divisibility | AMC-10A, 2003 | Problem 15

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

There are total number of integers are \(100\).and numer of integers divisible by \(2\) is \(\frac{100}{2}\)=\(50\). Now we have to find out divisible by \(2\) and not divisible by \(3\). so at first we have to find out the numbers of integers which are divisible by \(2\) and \(3\) both.......

can you finish the problem........

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

Therefore numbers of integers which are divisible by \(6\)=\(\frac{100}{6}=16\) (between \(1\) & \( 100\))

can you finish the problem........

Therefore the number of integers which are divisible by \(2\) and not divisible by \(3\)= \(50 - 16=34\).

So require probability= \(\frac{34}{100}=\frac{17}{50}\)

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