The positive integer k has exactly two positive prime factors, 3 and 7

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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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

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)

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?

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....?

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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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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K > 0
K = \(3^a\)*\(7^b\)
(a+1)(b+1)=6
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
_________________

Bunuel,

Are there other good questions to practice this concept you could point me to?

Thanks a lot,

Hadrien

5. Divisibility/Multiples/Factors

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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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Given \(K = 3^x * 7^y\) & \((x+1)(y+1) = 6\)

Hence x can take 2 values to give integer values of y
\(x = 2\), we get \(y = 1\) & hence \(K = 3^2 * 7\)
\(x = 1\), we get \(y = 2\) & hence \(K = 3 * 7^2\)

Statement 1:
\(3^2\) is a factor of \(K\)
Hence \(y = 1\) & \(K = 3^2 * 7\)

Statement 1 alone is Sufficient.

Statement 2:
\(7^2\) is NOT a factor of \(K\)
Hence \(x = 2\) & \(y = 1\)
\(K = 3^2 * 7\)

Statement 2 alone is Sufficient.

Answer D.


Thanks,
GyM
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Since there are very few options (k has only 6 positive factors and 2 prime factors), we can just write them all out.
This is an Alternative approach.

First, we know k has 1,3,7,21 and k as distinct factors, meaning that we're missing exactly one. (k can't be 21 because then 1,3,7,21 would be all the positive factors...)

(1) this must be our last positive factor and is therefore sufficient. (In particular, it means that k is the LCM of 1,3,7,9, and 21, which is 9*7 = 63)
Sufficient.

(2) this is equivalent to statement (1): if we can't increase the power of 7, we must increase the power of 3 (because these are the only prime factors), so 3^2 is a factor of k.
Sufficient

(D) is our answer.

Solution:

Since k has only two prime factors, 3 and 7, k = 3^m x 7^n where m and n are positive integers. In this case, the number of factors of k is (m + 1)(n + 1). Since we are given that k has 6 factors, we have:

(m + 1)(n + 1) = 6

Since both m and n are integers greater than or equal to 1, we see that m = 1 and n = 2 OR m = 2 and n = 1.

Statement One Only:

3^2 is a factor of k

We see that m = 2. So n = 1 and therefore, k = 3^2 x 7^1 = 63. Statement one alone is sufficient.

Statement Two Only:

7^2 is NOT a factor of k.

We see that n ≠ 2. So n must be 1 and m must be 2. Again we have k = 3^2 x 7^1 = 63. Statement two alone is sufficient.

Answer: D
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This question is a great candidate for the “Break down the question statement” strategy.

Positive integer k has two positive prime factors, 3 and 7. This means that k is definitely a composite number since it has more than 2 factors (remember that 1 and k are already factors of k, so there are more than 2 factors for sure).

For any composite number N which can be written as N = \(a^x * b^y * c^z \)* ……………, the number of positive factors of N is given by the expression (x+1) (y+1) (z+1)……….

As per the question statement, k can be written as k = \(3^x * 7^y\). Therefore, the number of factors for k = (x+1) (y+1). The questions tells us that k has a total of 6 positive factors.
So, (x+1) (y+1) = 6.

Note that x and y are positive integers, therefore, it’s easier to deal with the equation by plugging in values for x and y than multiplying the terms in the bracket.

If (x+1) = 3, (y+1) = 2; so x = 2 and y = 1 and k = \(3^2 * 7^1\).

If (x+1) = 2, (y+1) = 3; so x = 1 and y = 2 and k = \(3^1 * 7^2\).
By breaking down the question, we understand that there can be only two possible values for k. Let’s now use the statements to figure out which one it is.

From statement I alone, \(3^2\) is a factor of k. This means k = \(3^2 * 7^1\), since a factor cannot be bigger than the number it divides (\(3^2\) cannot divide \(3^1\))
Statement I alone is sufficient. Answer options B, C and E can be eliminated. Possible answer options are A or D.

From statement II alone, \(7^2\) is not a factor of k. This again means k = \(3^2 * 7^1\).
Statement II alone is sufficient. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
Aravind B T

We can use the theory of finding "number of factors" as posted by bunuel earlier. Hence, by reversly using the theory we know that K can be either (3^2 * 7) or (3*7^2).

After doing this work, we can easily use statement I and II to find the answers.
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