7Amulya
Hi Vaibhav708,
Please elaborate step by step. I am not able to figure out how you did this.
Vaibhav708
There is an easier way to find prime numbers. First find the square root of nearest and lowest perfect square. If none of the prime numbers till that square root divides the given number, then it is a prime number.
Nearest root of 117 is 10. 117 is divisible by 3. Not a prime no.
Nearest root of 133 is 11. 133 is divisible by 7. Not a prime no.
Nearest root of 189 is 13. 189 is divisible by 3. Not a prime no.
Nearest root of 209 is 14. 209 is divisible by 11. Not a prime no.
Nearest root 223 is 14. No prime lower than and equal to 14 divides 223.
Therefore 223 is a prime number.(E)
For option A. and C., the sum of the digits is multiple of 3 => so the numbers have to be divisible by 3, they cannot be prime.
The rest can be checked by the following method: Verifying the primality of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all prime integers
smaller than \(\sqrt{n}\) (if \(\sqrt{n}=integer\) then n is obviously not a prime), thereby checking whether \(n\) is a multiple of \(prime<\sqrt{n}\).
Verifying the primality of \(133\): \(\sqrt{133}\) is little less than \(12\). We should check \(133\) on divisibility by prime numbers from 2 to 12: \(133\) is divisible by \(7\), hence \(133\) is not prime.
Verifying the primality of \(209\): \(\sqrt{149}\) is little more than \(14\). We should check \(209\) on divisibility by prime numbers from 2 to 14. \(209\) is divisible by \(11\), hence \(209\) is not prime.
Option E. has to be the answer.Verifying the primality of \(223\): \(\sqrt{223}\) is little less than \(15\). We should check \(223\) on divisibility primes from 2 to 15. \(223\) is not divisible by any of the integers from \(2\) to \(15\), hence \(223\) is prime.