jjack0310 wrote:
From the stem, we know that K's factors are 1, 3, 7, 21 (3*7), __, and K.
1) This tells us there are two factors of 3, so 9 is also a factor of K. K's factors are 1, 3, 7, 9, 21, and K. Since there are two 3's and a 7 in K's factors, then 3*3*7 = 63 is also a factor.
Therefore K's factors are 1, 3, 7, 9, 21, 63.
SUFFICIENT
2) If there are not 2 7's in K's factors, and there are exactly 6 factors total, there must be two factors of 3. Otherwise, if we were to use a non-prime factor, then K would have more than 6 factors. (Remember 'K' has exactly two positive prime factors)
Therefore, K's factors are 1, 3, 7, 9, 21, 63.
SUFFICIENT
Answer is D.
The bold part is what I do not understand. I am sorry, but I dont get the factors part where it says "there are two 3's and a 7 in K's factors".
Can someone please explain why is this the case? What allows us to say this? I mean what allows us to say two 3's and a 7? 9 is 3^2, 21 is 3*7, but....?
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:The positive integer k has exactly two positive prime factors, 3 and 7. If k has a total of 6 positive factors, including 1 and k, what is the value of K?"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) 3^2 is a factor of k --> we have the second case, hence \(k=3^2*7^1=9*7\). Sufficient.
(2) 7^2 is NOT a factor of k --> we have the second case, hence \(k=3^2*7^1=9*7\). Sufficient.
Answer: D.
Hope it's clear.
Thank you much Bunuel.
Just one last question, and the reason that we are not acounting for the case when m = 0, and n = 5 is because 3^0 or 7^0 would be 1, and in that case, 3 is not a prime factor of k. Correct?