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The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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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? (1) 3^2 is a factor of k (2) 7^2 is NOT a factor of k Attachment:
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I searched thru 67 pages using keywords, but I did not find this question asked, I think this could be a newly added question in the gmatprep software. somewhat of a tricky wording question, especially when time is running short. oa is d. correction: oa is D.
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Originally posted by gmatnub on 27 Feb 2008, 20:33.
Last edited by Bunuel on 04 Jan 2018, 07:54, edited 3 times in total.
Renamed the topic, edited the question and added the OA.




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The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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31 Dec 2013, 04:16
jjack0310 wrote: samiam7 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 nonprime 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.
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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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26 Sep 2009, 10:39
Positive integer 'K' has exactly two positive prime factors, 3 and 7. If 'K' has a total of 6 factors, including 1 and 'K', what is the value of 'K'?
(1) 3^2 is a factor of 'K'
(2) 7^2 is not a factor of 'K'.
Soln: Since k has two positive prime factors k = 3^a * 7^b k has a total of 6 factors meaning (a+1) * (b+1) = 6 this can be either (a+1) * (b+1) = 1 * 6 or (a+1) * (b+1) = 2 * 3
1 * 6 is not possible because one of the factors will become 0. In tat case k will have just one prime factor. Hence the only option is 2 * 3 So when a = 2, b = 1 and when a = 1, b = 2 thus k can be either 3^2 * 7^1 or 3^1 * 7^2
Now considering statement 1 alone, 3^2 is a factor of 'K'. This will be true only when k = 3^2 * 7^1 Thus statement 1 alone is sufficient
Now considering statement 2 alone, 7^2 is not a factor of 'K'. This will be true only when k = 3^2 * 7^1 Thus statement 2 alone is sufficient
Hence D




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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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27 Feb 2008, 21:34
Based on stem, the 6 factors of k are 1,3,7,21, x and k . where 7 < x < k.
If statement (1) is used, the factors are, 1, 3, 7, 9, 21, k. k = 63. sufficient Since stem says 3,7 are the only prime factors, x has to be 3^2 since x cannot be 7^2.  sufficient
Answer (C)



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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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28 Feb 2008, 07:19
gmatnub wrote: Gmatprep DS: 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?
1) 3^2 is a factor of k
2) 7^2 is NOT a factor of k
I searched thru 67 pages using keywords, but I did not find this question asked, I think this could be a newly added question in the gmatprep software.
somewhat of a tricky wording question, especially when time is running short. oa is a. K has 6 factors: 1,3,7,21,X,K (different factors) Essentially we need to find X then we will know K. 1: X must be 9. b/c K has two 3's as factors. 2: if 7^2 is not a factor of K then X cannot be 49. Since we only have 3 and 7 as prime factors, 3 must be the other factor and X would be 9. I get D Im not sure why OA is A... =(



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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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16 Feb 2009, 22:55
Good question. I have a different way of solving this.
Let P1 = Power of first factor Let P2 = Power of second factor The number of factors can be found using the equation (P1 + 1)(P2 + 1). This is a rule, I didn't come up with this. Therefore here we have: 2*3 or 3*2, both equal 6.
statement 1: says that 3*2 is out, therefore sufficient statement 2: says that 3*2 is out, therefore sufficient.
note that we cannot use 6*1, because then we have a 7^0 or a 3^0, which is not the case here.
Answer D.
What do you think?



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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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19 Feb 2009, 11:07
x1050us wrote: Based on stem, the 6 factors of k are 1,3,7,21, x and k . where 7 < x < k.
If statement (1) is used, the factors are, 1, 3, 7, 9, 21, k. k = 63. sufficient Since stem says 3,7 are the only prime factors, x has to be 3^2 since x cannot be 7^2.  sufficient
Answer (C) I don't understand why k=63, why can't it be 27 (due to 3 x 9)??



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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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19 Feb 2009, 12:11
DaveGG wrote: x1050us wrote: Based on stem, the 6 factors of k are 1,3,7,21, x and k . where 7 < x < k.
If statement (1) is used, the factors are, 1, 3, 7, 9, 21, k. k = 63. sufficient Since stem says 3,7 are the only prime factors, x has to be 3^2 since x cannot be 7^2.  sufficient
Answer (C) I don't understand why k=63, why can't it be 27 (due to 3 x 9)?? In that case, k would have 3^3 as factor. If so, the k would have more than 6 factors as under: 1, 3, 7, 9, 21, 27, 42, 63, and 189 gmatnub wrote: Gmatprep DS: 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?
1) 3^2 is a factor of k 2) 7^2 is NOT a factor of k We need one more either 3 or 7 to have 6 +ve factors of k. a: 3^2 makes 6 +ve factors. b. if there is no 7^2 as a factor of k, then it also makes sure that 3^3 is a factor of k.
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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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23 Sep 2009, 09:24
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 nonprime 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.



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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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16 Nov 2013, 14:43
gmatnub wrote: 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?
(1) 3^2 is a factor of k (2) 7^2 is NOT a factor of k
The solutions that try to name each factor are dangerous because one can always run the risk to overlook one or two factors. Oddly enough, I feel that the best way to approach this problem is through " combinatories"! It is just a matter of seeing that the total number of factors in K (6 as mentioned in the stem) is the product of the "group of possible factors including 3" and "the group of possible factors including 7". Statement one is sufficient: As per the statement, the group of possible factors including 3 is 3 (0, 1 or 2 times)  therefore 3 possibilities. We do know that total number of factors of K is 6, so the group of possible factors including 7 has to be two  when 7 appears 0 or 1 time. So group of three  three elements (0,1 or 2) times group of 7  two elements (0 or 1) equals 6! Statement two is also sufficient: The only possible factors of K is 6, so either "the group of factors including 7" is two (7^1) or three (7^2) possibilities. The statement rules out the later, that leaves you with two possibilities for "the group of factors including 7".



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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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30 Dec 2013, 19:14
samiam7 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 nonprime 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....?



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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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01 Jan 2014, 09:58
Bunuel wrote: 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 nonprime 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?



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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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02 Jan 2014, 05:19
jjack0310 wrote: Bunuel wrote: 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 nonprime 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? Absolutely, m and n must be greater than zero because if they are not then 3 and 7 are not the factors of k.
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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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08 Jan 2017, 22:39
Given : \(k=3^n * 7^m\) (1), where m & n are powers of prime factors, 3 and 7. also we know that k has a total of 6 positive factors, including 1 and k
this can be represented as (n+1)(m+1)=6 (2)
statement (1) : 3^2 is a factor of k n=2 , substitute in (2) , we get m=1 put n=2 & m=1 in (1) , we get \(k=3^2 * 7^1\) >> k=63 >> sufficient.
statement (2) : \(7^2\) is NOT a factor of k as from the question stem , we know that 7 is among the prime factors of k, hence ,the minimum power of 7 is 1. therefore m=1 , substitute in (2) , we get n=2 put n=2 & m=1 in (1) , we get \(k=3^2 * 7^1\) >> k=63 >> sufficient.
Ans : D



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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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29 Jan 2017, 17:24
statement (1) if 3^2 is a factor of k, then so is 3^1. therefore, we already have four factors: 1, 3^1, 3^2, and 7. but we also know that (3^1)(7) and (3^2)(7) must be factors, since 3^2 and 7 are both part of the prime factorization of k. that's already six factors, so we're done: k must be (3^2)(7). if it were any bigger, then there would be more than these six factors. sufficient. statement (2) if 7 is a factor of k, but 7^2 isn't, then the prime factorization of k contains EXACTLY one 7. therefore, we need to find out how many 3's will produce six factors when paired with exactly one 7. in fact, it's data sufficiency, so we don't even have to find this number; all we have to do is realize that adding more 3's will always increase the number of factors, so, there must be exactly one number of 3's that will produce the correct number of factors. (as already noted above, that's two 3's, or 3^2.) sufficient. Hence D.
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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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12 Aug 2017, 12:12
gmatnub wrote: 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 > 0 K = \(3^a\)*\(7^b\) (a+1)(b+1)=6 K = ? Quote: (1) 3^2 is a factor of k (2) 7^2 is NOT a factor of k
1) \(3^2\) is a factor of K => (2+1)(b+1)=6 => b = 1 We can find the value of K. Sufficient. 2) \(7^2\) is NOT a factor of K (a+1)(b+1)=6 => b≠ 2 ; b can only be 1. => a = 2, b = 1 We can find the value of K. Sufficient. D is the answer. Great Official Question
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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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02 Dec 2017, 08:26
Bunuel, Are there other good questions to practice this concept you could point me to? Thanks a lot, Hadrien
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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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Fantastic thank you. I also found your incredible thread on tips and hints for specific quant topics! Looks like I have a full day worth of material ahead of me. Thanks! https://gmatclub.com/forum/tipsandhin ... l#p1379270Bunuel wrote: Hadrienlbb wrote: Bunuel, Are there other good questions to practice this concept you could point me to? Thanks a lot, Hadrien 5. Divisibility/Multiples/Factors For other subjects: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative Megathread
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Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]
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11 Dec 2017, 18:36
gmatnub wrote: 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?
(1) 3^2 is a factor of k (2) 7^2 is NOT a factor of k We are given that K has two positive prime factors, 3 and 7, and that K has a total of 6 factors including 1 and K. Thus, we know that the factors of K include 1, 3, 7, 21, and K. We must determine the value of K. Statement One Alone: 3^2 is a factor of K. Let’s list the factors of K: 1, 3, 7, 21, K, and 3^2 = 9 Since 3 and 7 are factors of K, 3 x 7 = 21 must also be a factor of K. Similarly, since 9 and 7 are both factors of K (and they are relatively prime), 9 x 7 = 63 must also be a factor of K. Since we already have 6 factors, K must equal 63. Statement one alone is sufficient to answer the question. Statement Two Alone: 7^2 is not a factor of K. If 7 is a factor of K but 7^2 is not a factor of K, then 3^2 = 9 must be a factor of K (otherwise, K has only 4 factors, namely 1, 3, 7, and 21). If 9 is a factor of K, then the list of factors of K is 1, 3, 7, 9, 21, 63. Therefore, K = 63. Statement two alone is also sufficient to answer the question. Alternative solution: We need to determine the value of K. We are given that K has two positive prime factors, 3 and 7. Therefore, the prime factorization of K must be K = 3^m x 7^n for some positive integers m and n greater than 1. Recall that the total number of factors of a number can be obtained by multiplying the numbers resulting from adding 1 to the exponents in the prime factorization. Thus, the total number of factors of K is (m + 1) x (n + 1). Since we are given that K has a total of 6 factors, (m + 1) x (n + 1) = 6. Since m and n are both greater than 1, either m = 2 and n = 1 OR m = 1 and n = 2. Statement One Alone: 3^2 is a factor of K. This tells us that m = 2, so n = 1. Therefore, K = 3^2 x 7^1 = 63. Statement one alone is sufficient to answer the question. Statement Two Alone: 7^2 is not a factor of K. This tells us that n ≠ 2, so n = 1, and thus m = 2. Therefore, K = 3^2 x 7^1 = 63. Statement two alone is sufficient to answer the question. Answer: D
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Re: The positive integer k has exactly two positive prime factors, 3 and 7
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11 Dec 2017, 18:36



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