banksy wrote:
187. Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 68
(D) 79
(E) 88
Any prime number more than 3 can be expressed as \(p=6n+1\) or\(p=6n+5\) (\(p=6n-1\)), where n is an integer >0 (check this:
primality-check-108425.html).
So the sum of two primes more than 3 can yield the following remainders upon division by 6:
0 - if two primes are of a type \(p=6n+1\) and \(p=6n+5\);
2 - if both primes are of a type \(p=6n+1\);
4 - if both primes are of a type \(p=6n+5\);
Now, we are looking for the choice which is not a prime+2, or prime+3 or has a remainder other than 0, 2, or 4 upon division by 6.
(A) 19 --> 19-2=17=prime;
(B) 45 --> 45-2=43=prime;
(C) 68 --> yields a remainder of 2 upon division by 6 so theoretically can be the sum of two primes (and it is 61+7=68);
(D) 79 --> 79-2 is not a prime, 79-3 is not a prime and also 79 yields a remainder of 1 upon division by 6, so it can not be the sum of two primes;
(E) 88 --> 71+17=88.
Answer: D.
Of course the above can be done much easier by just subtracting the primes starting from 2 from the answer choices and seeing whether the result is also a prime.
Awesome explanation based on the crucial theory that any prime number >3 can be expressed in 6n+1 or 6n-1 format. I did not know this hence ended up taking an alternative logic. Where can I find more of such theorems on number theory.