Try this beautiful problem from AMC 10A, 2003 based on Probability in 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
But try the problem first...
Answer: \(\frac{17}{50}\)
AMC-10A (2003) Problem 15
Pre College Mathematics
First hint
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........
Second Hint
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........
Final Step
Therefore the number of integers which are divisible by \(2\) and not divisible by \(3\)= \(50 - 16=34\).
So require probability= \(\frac{34}{100}=\frac{17}{50}\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
Try this beautiful problem from AMC 10A, 2003 based on Probability in 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
But try the problem first...
Answer: \(\frac{17}{50}\)
AMC-10A (2003) Problem 15
Pre College Mathematics
First hint
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........
Second Hint
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........
Final Step
Therefore the number of integers which are divisible by \(2\) and not divisible by \(3\)= \(50 - 16=34\).
So require probability= \(\frac{34}{100}=\frac{17}{50}\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
