If a, f, and c are consecutive even integers and a < f < c, all of the following must be divisible by 4 EXCEPT	Pick
A) a + c B) f + c C) ac D) (bc)/2 E) (abc)/4
I did get the correct answer within 80 seconds and it was not by luck either. I did not pick numbers but just used the concept stated by Ian.   The 3 numbers can be written as   a, (a + 2)(a + 4).   If 'a' is divisible by 4, then even 'c' or 'a + 4' is divisible by 4. However, is 'b' is divisible by 4, then both 'a' and 'a + 4' are still divisible by 2.   A - (a + c) = a + (a + 4) = 2a + 4 = 2(a + 2) = 2b. 2b will always be divisible by 4 even if 'b' is not divisible by 4. Reason: 'b' already has a prime factorization of at least a '2'. Hence '2b' has two 2s.   C - ac = a(a+4). If, as stated above, one of them is divisible by 4, then the product is divisible. If both of them are not divisible by 4, then the product is still divisible by 4 because of the presence of two 2s again in the prime factorization.   D - bc/2 = (a + 2)(a + 4)/2. Either b or c is divisible by 2. Hence, if we assume that b is divisible by 2 and not divisible by 4, then it leaves us just one possibility. Is c divisible by 4? It has to be because c is the next consecutive even integer.   E - abc/4 = a(a + 2)(a + 4)/4. One of these integers is divisible by 4 already. If we again assume 'b' to be that integer divisible by 4, then we are left with the question - Is a(a + 4) divisible by 4? This is the same as option C.   B - f + c = (a + 2) + (a + 4) = 2a + 6 = 2(a + 3). (a + 3) will never be divisible by 2 because it is an odd integer. Hence, 2(a + 3), although divisible by 2, will not be divisible by 4 because it has just one 2 in its prime factorization.   As a whole, whether you choose numbers (2, 46 being the easiest) or solve conceptually, the answer is still easily obtainable within 2 minutes.f + c
<result>B</result>
If a, f, and c are consecutive even integers and a < f < c, all of the following must be divisible by 4 EXCEPT