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

But try the problem first…

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

AMC-10A (2003) Problem 15

Pre College Mathematics

## Try with Hints

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

## Other useful links

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