Try this beautiful problem from Probability based on divisibility.
What is the probability that an integer in the set \({1,2,3,…,100}\) is divisible by \(2\) and not divisible by \(3\)?
Number system
Probability
divisibility
Answer: \(\frac{17}{50}\)
AMC-10A (2003) Problem 15
Pre College Mathematics
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 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 wcich 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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