Answer
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Hint: For identifying the numbers which are prime or not we will use the divisibility rule for answering it. The number which is divisible only $ 1 $ and itself is termed as the prime number. Also, we know that the number which is divisible by $ 1 $ and itself and also by some other numbers is known to be a composite number. From this, we can easily find those numbers now.
Complete step-by-step answer:
Here, we already have an understanding of the prime and composite numbers from the hint. And we should also know that the composite numbers will not be a prime number. So we will find the composite number and then the left numbers will be a prime number.
For finding this we will use the divisibility rule.
As we know that the numbers which are divisible by $ 2 $ are even numbers. So all divisible numbers except one number which is $ 2 $ itself will be a composite number.
Hence, we have $ 62,64,.....96,98 $ .
Now we will check for divisibility of $ 5 $ . So for this, we know that the numbers end with $ 0 $ or $ 5 $ comes under this, therefore, the numbers will be $ 65,70,75.....85,90,95 $
So after doing these two checks we have the numbers left are
$ 61,63,67,69,71,73,77,79,81,83,87,89,91,93,97,99 $
Now we will check the divisibility of these numbers by $ 3 $ and $ 7 $
As we know that the number which is divisible by $ 3 $ if their sum of digits is divisible by $ 3 $
So we have the numbers left are
$ 61,67,71,73,77,79,83,89,91,97 $
Now we will check the divisibility of these numbers by $ 7 $
As we know that the number which is divisible by $ 7 $ if we twice the units place of the digit which will be subtracted from the rest numbers will be divisible by $ 7 $
i.e. $ 9 - \left( {1 \times 2} \right) = 9 - 2 = 7 $ , hence the number $ 91 $ will be divisible by $ 7 $ . Similarly $ 77 $ will be divisible by $ 7 $ .
So we will remove these numbers also.
The remaining prime numbers will be $ 61,67,71,73,79,83,89,97 $ .
So, the correct answer is “ $ 61,67,71,73,79,83,89,97 $ ”.
Note: So we have successfully got the prime numbers between those numbers. While solving such a question we should always be careful and read correctly whether it is saying between them or from this to this. The divisibility test gives us a favor to eliminate many numbers from that particular range quickly. That’s why it is preferable.
Complete step-by-step answer:
Here, we already have an understanding of the prime and composite numbers from the hint. And we should also know that the composite numbers will not be a prime number. So we will find the composite number and then the left numbers will be a prime number.
For finding this we will use the divisibility rule.
As we know that the numbers which are divisible by $ 2 $ are even numbers. So all divisible numbers except one number which is $ 2 $ itself will be a composite number.
Hence, we have $ 62,64,.....96,98 $ .
Now we will check for divisibility of $ 5 $ . So for this, we know that the numbers end with $ 0 $ or $ 5 $ comes under this, therefore, the numbers will be $ 65,70,75.....85,90,95 $
So after doing these two checks we have the numbers left are
$ 61,63,67,69,71,73,77,79,81,83,87,89,91,93,97,99 $
Now we will check the divisibility of these numbers by $ 3 $ and $ 7 $
As we know that the number which is divisible by $ 3 $ if their sum of digits is divisible by $ 3 $
So we have the numbers left are
$ 61,67,71,73,77,79,83,89,91,97 $
Now we will check the divisibility of these numbers by $ 7 $
As we know that the number which is divisible by $ 7 $ if we twice the units place of the digit which will be subtracted from the rest numbers will be divisible by $ 7 $
i.e. $ 9 - \left( {1 \times 2} \right) = 9 - 2 = 7 $ , hence the number $ 91 $ will be divisible by $ 7 $ . Similarly $ 77 $ will be divisible by $ 7 $ .
So we will remove these numbers also.
The remaining prime numbers will be $ 61,67,71,73,79,83,89,97 $ .
So, the correct answer is “ $ 61,67,71,73,79,83,89,97 $ ”.
Note: So we have successfully got the prime numbers between those numbers. While solving such a question we should always be careful and read correctly whether it is saying between them or from this to this. The divisibility test gives us a favor to eliminate many numbers from that particular range quickly. That’s why it is preferable.
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