
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Answer
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Hint: Firstly, we will list all the natural numbers between 102 and 113, Then we check for the divisibility of 2,3,5,7,11, etc. for the numbers listed. We know that prime numbers are only divisible by one and itself. If any of the numbers listed are divisible by any other number, we won’t consider it as a prime number. After finding the prime numbers, we will find the ratio of the total number of prime numbers to that of the total number of natural numbers between 102 and 113.
Complete step-by-step answer:
The given numbers are 102 and 113
We know that natural numbers can be considered as the numbers which are greater than 0. Hence the natural numbers between 102 and 113 can be written as:
102, 103, 104, 105, 106, 107,108, 109, 110, 111,112,113
Hence there are a total 12 natural numbers between 102 and 113.
We also know that prime numbers are the numbers that can be divisible by 1 and itself. So, we will check for the numbers listed above, which are divisible by any other number.
If the number is ending with 0, 2, 4, 6, 8, then it must be divisible by 2. Hence we can say that the numbers 102, 104, 106, 108, 110, 112 cannot be a prime number because they are divisible by 2.
If the sum of the digits of any number is divisible by 3 then it must be divisible by 3. Hence we can say that the numbers 105 and 111 cannot be prime because they are divisible by 3.
Now, we will check for divisibility of 5, 7 and 11. None of the numbers left is divisible by them. Hence we can say that 103, 107, 109, 113 are prime numbers.
Prime numbers between 102 and 113 can be written as:
103, 107, 109, 113
Therefore, there are 4 prime numbers between 102 to 113.
To find the fraction of the prime numbers, we will find the ratio of the total numbers of prime numbers and total number of natural numbers. This can be expressed as:
$\begin{array}{c}
{\rm{Required fraction}} = \dfrac{4}{{12}}\\
{\rm{Required fraction}} = \dfrac{1}{3}
\end{array}$
Note: We should have prior knowledge about the prime numbers. In this question, we are checking for the divisibility of 2, 3, 5, 7, 11, etc. Hence, we should know the divisibility test of these numbers. There is a certain condition for the divisibility test of any number.
Complete step-by-step answer:
The given numbers are 102 and 113
We know that natural numbers can be considered as the numbers which are greater than 0. Hence the natural numbers between 102 and 113 can be written as:
102, 103, 104, 105, 106, 107,108, 109, 110, 111,112,113
Hence there are a total 12 natural numbers between 102 and 113.
We also know that prime numbers are the numbers that can be divisible by 1 and itself. So, we will check for the numbers listed above, which are divisible by any other number.
If the number is ending with 0, 2, 4, 6, 8, then it must be divisible by 2. Hence we can say that the numbers 102, 104, 106, 108, 110, 112 cannot be a prime number because they are divisible by 2.
If the sum of the digits of any number is divisible by 3 then it must be divisible by 3. Hence we can say that the numbers 105 and 111 cannot be prime because they are divisible by 3.
Now, we will check for divisibility of 5, 7 and 11. None of the numbers left is divisible by them. Hence we can say that 103, 107, 109, 113 are prime numbers.
Prime numbers between 102 and 113 can be written as:
103, 107, 109, 113
Therefore, there are 4 prime numbers between 102 to 113.
To find the fraction of the prime numbers, we will find the ratio of the total numbers of prime numbers and total number of natural numbers. This can be expressed as:
$\begin{array}{c}
{\rm{Required fraction}} = \dfrac{4}{{12}}\\
{\rm{Required fraction}} = \dfrac{1}{3}
\end{array}$
Note: We should have prior knowledge about the prime numbers. In this question, we are checking for the divisibility of 2, 3, 5, 7, 11, etc. Hence, we should know the divisibility test of these numbers. There is a certain condition for the divisibility test of any number.
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