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Number Properties

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by ela07mjt » Thu Feb 07, 2013 5:09 am
Need help with this!

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,
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what is the value of k ?
(1) 3^2 is a factor of k.
2) 7^2 is not a factor of k.
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Source: — Data Sufficiency |

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by Ian Stewart » Thu Feb 07, 2013 5:25 am
From the stem, we know the prime factorization of k must look like (3^a)(7^b), where a and b are positive integers. Now, to count how many divisors a number has, we add 1 to each exponent in its prime factorization and multiply. So k must have (a+1)(b+1) divisors. Since k has 6 divisors, (a+1)(b+1) = 6. Now we're multiplying two integers greater than 1 (since a and b are greater than 0) and getting 6 as a result, so we must be multiplying 2 and 3. So either a=1 and b=2, or a=2 and b=1.

Statement 1 guarantees that a is at least 2, so the only possibility is that a=2 and b=1. Statement 2 guarantees that b < 2, so the only possibility is that b=1 and a=2. So each Statement is sufficient alone, and the answer is D.
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by GMATGuruNY » Thu Feb 07, 2013 5:53 am
ela07mjt wrote:Need help with this!

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.
If k = 3*7 = 21, then k has the following positive factors:
1*21
3*7
A total of 4 positive factors.

Since k has 6 positive factors -- and its prime-factorization can be composed ONLY OF 3's and 7's -- there are only TWO POSSIBLE CASES:

Case 1: k = 1*3*3*7 = 63, in which case k has the following positive factors:
1*63
3*21
7*9

Case 2: k = 1*3*7*7 = 147, in which case k has the following positive factors:
1*147
3*49
7*21

If the prime-factorization of k includes any more 3's or 7's, the total number of positive factors will increase beyond 6, violating the condition that k has exactly 6 positive factors.
Thus, only Case 1 or Case 2 is possible.

Statement 1: 3² is a factor of k
Thus, k = Case 1 = 63.
SUFFICIENT.

Statement 2: 7² is not a factor of k
Since Case 2 is not possible, k = Case 1 = 63.
SUFFICIENT.

The correct answer is D.
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by cking6178 » Thu Feb 07, 2013 6:33 am
GMATGuruNY wrote:
ela07mjt wrote:Need help with this!

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.
If k = 3*7 = 21, then k has the following positive factors:
1*21
3*7
A total of 4 positive factors.

Since k has 6 positive factors -- and its prime-factorization can be composed ONLY OF 3's and 7's -- there are only TWO POSSIBLE CASES:

Case 1: k = 1*3*3*7 = 63, in which case k has the following positive factors:
1*63
3*21
7*9

Case 2: k = 1*3*7*7 = 147, in which case k has the following positive factors:
1*147
3*49
7*21

If the prime-factorization of k includes any more 3's or 7's, the total number of positive factors will increase beyond 6, violating the condition that k has exactly 6 positive factors.
Thus, only Case 1 or Case 2 is possible.

Statement 1: 3² is a factor of k
Thus, k = Case 1 = 63.
SUFFICIENT.

Statement 2: 7² is not a factor of k
Since Case 2 is not possible, k = Case 1 = 63.
SUFFICIENT.

The correct answer is D.
Thanks Guru!! DS is going to be the death of me...I factored K out and solved for the 63 case, but left out the 147 case! Test is Saturday, need to get DS under control.
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