If the prime factorization of the integer q can be expressed as a^2xb^xc^3x−1, where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q? Pick one.
A) 3j + 4, where j is a positive integer B) 5k + 5, where k is a positive integer C) 6l + 2, where l is a positive integer D) 9m + 7, where m is a positive integer E) 10n + 1, where n is a positive integer
Straight forward question if you know the formula for number of factors of an integer q=ax∗by∗czq=ax∗by∗cz , where x,y,z≥0 x,y,z≥0 and are integers = (x+1)(y+1)(z+1) including 1 and integer q.   Now given, q= a^2xb^xc^3x−1, thus based on the above formula, the number of factors of q including 1 and q = (2x+1)(x+1)(3x-1+1) = 3x(2x+1)(x+1). So we know that the number will be of the form 3x(2x+1)(x+1) , will be an integer and will be true for ALL X. Thus substitute x=1, we get number of factors = 3*3*2= 
<gadget id="calculator">3*3*2</gadget>
<output>18</output>
18 (this eliminates A, as 3 is a factor of 18 and thus can not leave 4 or (4-3=1) as the remainder. Similarly, eliminate C and D).   Now substitute x=2, we get number of factors = 6*5*3= 
<gadget id="calculator">6*5*3</gadget>
<output>90</output>
90 and this eliminates E as 10 is a factor of 90 and will leave a remainder of 0. Thus 5k + 5, where k is a positive integer is the only remaining answer choice   ANSWER:5k + 5, where k is a positive integer   .
<result>B</result>
If the prime factorization of the integer q can be expressed as a^2xb^xc^3x−1, where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?