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The positive integer k has exactly two positive prime factor [#permalink]
27 Feb 2008, 19:33

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Question Stats:

40% (01:54) correct
60% (01:19) wrong based on 287 sessions

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 6-7 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.

Re: Gmatprep DS: the positive integer k has exactly two [#permalink]
27 Feb 2008, 20:34

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

Re: Gmatprep DS: the positive integer k has exactly two [#permalink]
28 Feb 2008, 06:19

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 6-7 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.

Re: Gmatprep DS: the positive integer k has exactly two [#permalink]
16 Feb 2009, 21:55

2

This post received KUDOS

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.

Re: Gmatprep DS: the positive integer k has exactly two [#permalink]
19 Feb 2009, 10:07

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)??

Re: Gmatprep DS: the positive integer k has exactly two [#permalink]
19 Feb 2009, 11:11

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. _________________

Re: What is the value of K - Confusing one [#permalink]
23 Sep 2009, 08:24

6

This post received KUDOS

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.

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)

Re: What is the value of K - Confusing one [#permalink]
26 Sep 2009, 09:39

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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

Re: Positive integer 'K' has exactly two positive prime factors, [#permalink]
07 Nov 2013, 00:02

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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The positive integer k has exactly two positive prime factor [#permalink]
09 Nov 2013, 16:06

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

I searched thru 6-7 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.

Re: The positive integer k has exactly two positive prime factor [#permalink]
16 Nov 2013, 13:43

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".

Re: What is the value of K - Confusing one [#permalink]
30 Dec 2013, 18:14

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.

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)

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....?

Re: What is the value of K - Confusing one [#permalink]
31 Dec 2013, 03:16

9

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Expert's post

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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.

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)

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.

Re: What is the value of K - Confusing one [#permalink]
01 Jan 2014, 08:58

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.

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)

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?

Re: What is the value of K - Confusing one [#permalink]
02 Jan 2014, 04:19

Expert's post

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.

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)

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. _________________

Re: The positive integer k has exactly two positive prime factor [#permalink]
23 Feb 2015, 13:44

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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