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.