Try this beautiful problem from AMC 10A, 2003 based on Probability in Divisibility.

## 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 \(3\)?

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

**Key Concepts**

Number system

Probability

divisibility

## Check the Answer

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

AMC-10A (2003) Problem 15

Pre College Mathematics

## Try with Hints

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

## Other useful links

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