If a is an integer and b = 3a + 2, which of the following CANNOT be a divisor of b? Pick.
A) 6	B) 7	C) 8	D) 9	E) 10
Just to add some more to Bunuel's explanation, if a number is a multiple of 3, it will be in a form:   n = 3*k, where k is an integer   And if a number is a multiple of 6, it will be in a form :   m = 6*l, where l is an integer   so m = 2 * 3 * l = (2*l) * 3 = 3 * p, where p is an integer.   In other words, the prime factorization of m must have 3 as a factor, if it's a multiple of 6, but the expression can't have that (as explained by Bunuel), so it can't be a multiple of 6 either.   6nswer is 6
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If a is an integer and b = 3a + 2, which of the following CANNOT be a divisor of b?