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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 \(3\)?

- \(\frac {33}{100}\)
- \(\frac{1}{6}\)
- \(\frac{17}{50}\)
- \(\frac{1}{2}\)
- \(\frac{18}{25}\)

Number system

Probability

divisibility

But try the problem first...

Answer: \(\frac{17}{50}\)

Source

Suggested Reading

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= \(\frac{34}{100}=\frac{17}{50}\)

- https://www.cheenta.com/octahedron-problem-amc-10a-2006-problem-24/
- https://www.youtube.com/watch?v=XOrePzJWFiE

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