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Explore the Back-StoryTry 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

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

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

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

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

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

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Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

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