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The positive integer k has exactly two positive prime factor [#permalink]

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27 Feb 2008, 19:33

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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

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]

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27 Feb 2008, 20:34

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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]

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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]

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16 Feb 2009, 21:55

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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]

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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]

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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]

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23 Sep 2009, 08:24

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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]

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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: The positive integer k has exactly two positive prime factor [#permalink]

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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]

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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]

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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....?

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]

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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?

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]

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08 Jan 2017, 21:39

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.

Re: The positive integer k has exactly two positive prime factor [#permalink]

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29 Jan 2017, 16:24

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.
_________________

Re: The positive integer k has exactly two positive prime factors, 3 and 7 [#permalink]

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25 Jun 2017, 18:32

suntorytea 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 (3 squared) is a factor of k. (2) 7^2 (7 squared) is NOT a factor of k.

Help would be appreciated! Thanks.

Best regards, ST

Hint :- if k has 6 factors then k must be 3^2 *7 or k = 3* 7^2

Re: The positive integer k has exactly two positive prime factor [#permalink]

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25 Jun 2017, 20:57

suntorytea 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 (3 squared) is a factor of k. (2) 7^2 (7 squared) is NOT a factor of k.

Suggestion: First google or use gmatclub search to check whether question already exist on the forum or not If you have specific query then post on that existing thread. To tag people use "@"before username

Re: The positive integer k has exactly two positive prime factor [#permalink]

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25 Jun 2017, 21:02

Bunuel wrote:

suntorytea 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 (3 squared) is a factor of k. (2) 7^2 (7 squared) is NOT a factor of k.

Help would be appreciated! Thanks.

Best regards, ST

Merging topics. Please refer to the discussion above.

Thanks Bunuel for merging. By the time i could post my suggestion, topic was merged. But In general when we're previewing the answer, we get to see if someone has posted in meanwhile or not. but nothing like this in this case.
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