The positive integer k has exactly two positive prime

Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.


Back to the original question:
"k has exactly two positive prime factors 3 and 7" --> \(k=3^m*7^n\), where \(m=integer\geq{1}\) and \(n=integer\geq{1}\);
"k has a total of 6 positive factors including 1 and k" --> \((m+1)(n+1)=6\). Note here that neither \(m\) nor \(n\) can be more than 2 as in this case \((m+1)(n+1)\) will be more than 6.

So, there are only two values of \(k\) possible:
1. if \(m=1\) and \(n=2\) --> \(k=3^1*7^2=3*49\);
2. if \(m=2\) and \(n=1\) --> \(k=3^2*7^1=9*7\).


(1) 9 is a factor of k --> we have the second case, hence \(k=3^2*7^1=9*7\). Sufficient.

(2) 49 is a factor of k --> we have the first case, hence \(k=3^1*7^2=3*49\). Sufficient.

Answer: D.

BUT: in DS statements never contradict, so this cannot be real GMAT question. I guess one of the statements should be "x is NOT factor of k". In this case answer still would be D, but the question will be of GMAT type.

Hope it's clear.