With this question when I watched the video I didn’t understand the point Greg makes with we don’t need to try 11^2 and 13^2 because they’re greater? Basically my question is how do you solve this strategically (I knew to remove 111 because that adds up to 3 , i knew to look for even numbers but other than that I started guessing)

I think the point being made is that if a number is **not** a prime number, it must have a factor that’s less than the square root of the number.

Hence, since 11^{2} is greater than any of the options, there is no need to go that far.

A simple Solution Take all the numbers in this case 101,103,107,109,111 and 113, now take a prime number whose square is less than these numbers, in this case 11X11=121 is larger than all the given numbers, so wee need to check divisibility of these number with all the prime numbers that are less than 11, which are 2,3,5,7

Suppose a number 127 occurs in the option than we need to check the divisibility of all the option with prime number less than 13, which are 2,3,5,7,11

Thanks both for the response.