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Divisible by a positive integer -> factor No of factors for a number in the form (a^x)(b^y)(c^z) is given by (x+1)(y+1)(z+1)
20k = (2^2)(5^1)(k)
Stmt 1 says k is prime. so 20k = (2^2)(5^1)(k^1). Total # of factors is (2+1)(1+1)(1+1). So sufficient.
Stmy 2 says k = 7 so again Total # of factors is (2+1)(1+1)(1+1). So sufficient.
Hence answer is D, but that is not the OA. What am I missing?
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:
If k is a positive integer, then 20k is divisible by how many different positive integers?
\(20k=2^2*5*k\).
(1) k is prime. If \(k=2\), then \(20k=2^3*5\) --> the # of factors = \((3+1)(1+1)=8\) but if \(k=5\), then \(20k=2^2*5^2\) --> the # of factors = \((2+1)(2+1)=9\). Not sufficient.
(2) k = 7 --> \(20k=2^2*5*7\) --> the # of factors = \((2+1)(1+1)(1+1)=12\). Sufficient.
Re: If k is a positive integer, then 20k is divisible by
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22 Aug 2013, 04:21
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spjmanoli wrote:
If k is a positive integer, then 20k is divisible by how many different positive integers? 1. k is prime 2. k is 7
Divisible by a positive integer -> factor No of factors for a number in the form (a^x)(b^y)(c^z) is given by (x+1)(y+1)(z+1)
20k = (2^2)(5^1)(k)
Stmt 1 says k is prime. so 20k = (2^2)(5^1)(k^1). Total # of factors is (2+1)(1+1)(1+1). So sufficient.
Stmy 2 says k = 7 so again Total # of factors is (2+1)(1+1)(1+1). So sufficient.
Hence answer is D, but that is not the OA. What am I missing?
What you are missing in F.S 1, is that we don't know the value of k.
Scenario I: k=2, the total no of factors for 20k = \(2^2*5*2 = 2^3*5 = (3+1)*(1+1) = 8\)
Scenario II: k=3, the total no of factors for 20k = \(2^2*5*3 = (2+1)*(1+1)*(1+1) = 12.\) Hence, 2 different answers, thus, Insufficient.
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Re: If k is a positive integer, then 20k is divisible by how man
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24 Feb 2016, 11:43
Bunuel wrote:
Back to the original question:
If k is a positive integer, then 20k is divisible by how many different positive integers?
\(20k=2^2*5*k\).
(1) k is prime. If \(k=2\), then \(20k=2^3*5\) --> the # of factors = \((3+1)(1+1)=8\) but if \(k=5\), then \(20k=2^2*5^2\) --> the # of factors = \((2+1)(2+1)=9\). Not sufficient.
(2) k = 7 --> \(20k=2^2*5*7\) --> the # of factors = \((2+1)(1+1)(1+1)=12\). Sufficient.
Answer: B.
Hope it's clear.
Thanks, the outcome is clear but this method of picking random numbers to test with makes me very uneasy. If you get lucky and the 2 numbers you pick yield different results, then all is fine. But if they yield the same result, you don't know anything. Do you pick a 3rd candidate? A 4th?
Re: If k is a positive integer, then 20k is divisible by how man
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14 Mar 2016, 00:09
Here Statement ! is not sufficient as the prime may be 2 or 5 or any other prime. But statement 2 is sufficient as we Number of divisors now = 2*3*2= 12 Hence B
_________________
Divisible by a positive integer -> factor No of factors for a number in the form (a^x)(b^y)(c^z) is given by (x+1)(y+1)(z+1)
20k = (2^2)(5^1)(k)
Stmt 1 says k is prime. so 20k = (2^2)(5^1)(k^1). Total # of factors is (2+1)(1+1)(1+1). So sufficient.
Stmy 2 says k = 7 so again Total # of factors is (2+1)(1+1)(1+1). So sufficient.
Hence answer is D, but that is not the OA. What am I missing?
Very simple, no tricks here or anything. If we know the value of K then we obviously know the answer- though in the case of K being a prime number- what if K is 2? Then there is of course some overlap- do not forget 2 is a prime number.
Divisible by a positive integer -> factor No of factors for a number in the form (a^x)(b^y)(c^z) is given by (x+1)(y+1)(z+1)
20k = (2^2)(5^1)(k)
Stmt 1 says k is prime. so 20k = (2^2)(5^1)(k^1). Total # of factors is (2+1)(1+1)(1+1). So sufficient.
Stmy 2 says k = 7 so again Total # of factors is (2+1)(1+1)(1+1). So sufficient.
Hence answer is D, but that is not the OA. What am I missing?
20k = 2^2 * 5 * k
We already have 2 prime factors of 20k (2 and 5). The total number of factors depends on k.
(1) k is prime
k could be 2 or 5 or any other prime number. In each case, the number of factors would be different. If k = 2, \(20k = 2^3 * 5\) has 4*2 = 8 factors If k = 5, \(20k = 2^2 * 5^2\) has 3*3 = 9 factors If k is any other prime number such as 11, \(20k = 2^2 * 5 * 11\) has 3*2*2 = 12 factors Not sufficient
(2) k = 7 If k = 7, 20k = 2^2 * 5 * 7 Total number of factors = 3*2*2 = 12 Sufficient
Answer (B)
_________________
Karishma Veritas Prep GMAT Instructor
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Re: If k is a positive integer, then 20k is divisible by how man
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23 Nov 2018, 18:22
It’s easy to think that statement (1) is sufficient by thinking, “Well, I can figure out quickly that 20 has six factors (1, 2, 4, 5, 10, 20), and if k is a new prime number, then 20k will have six MORE factors than 20 has, because we can just take the new prime number and multiply it by each of the original six factors. For instance, if k were 3, then the factors of 20k would be each of the six factors of 20 PLUS six new ones - 3, 6, 12, 15, 30, 60. So statement (1) is sufficient.”
Statement (1) is a trap, however, because we aren’t told that k is a NEW prime number!!! It may be a prime that is ALREADY a factor of 20 – i.e., k could be 2 or 5. Suppose k were 2. If that were the case, then the factors of 20k would be 1, 2, 4, 5, 8, 10, 20 and 40 – i.e., we would be adding only two new factors, 8 and 40, to the original list of six. Statement (1) is therefore insufficient.
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Re: If k is a positive integer, then 20k is divisible by how man
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17 Mar 2019, 17:47
So everyone here agrees that the answer is "B." When I answered "B" on the OG online practice test the program said my answer was wrong and the answer is actually "D." Do we think the OG is wrong?
Re: If k is a positive integer, then 20k is divisible by how man
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18 Mar 2019, 04:49
5e3 wrote:
So everyone here agrees that the answer is "B." When I answered "B" on the OG online practice test the program said my answer was wrong and the answer is actually "D." Do we think the OG is wrong?
Was the question on the test slightly different? Can you put up a screenshot?
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Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Re: If k is a positive integer, then 20k is divisible by how man
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18 Mar 2019, 05:27
VeritasKarishma wrote:
5e3 wrote:
So everyone here agrees that the answer is "B." When I answered "B" on the OG online practice test the program said my answer was wrong and the answer is actually "D." Do we think the OG is wrong?
Was the question on the test slightly different? Can you put up a screenshot?
As soon as I posted this I saw that I read the correct answer wrong...ha! Thank you for the response, though!
gmatclubot
Re: If k is a positive integer, then 20k is divisible by how man
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18 Mar 2019, 05:27
close
59% (01:11) correct 
