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If k is a positive integer, then 20k is divisible by how man
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Updated on: 22 Aug 2013, 04:17
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If k is a positive integer, then 20k is divisible by how many different positive integers? (1) k is prime (2) k = 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?
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Originally posted by hellzangl on 22 Aug 2013, 04:14.
Last edited by Bunuel on 22 Aug 2013, 04:17, edited 1 time in total.
Edited the question.




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Re: If k is a positive integer, then 20k is divisible by how man
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22 Aug 2013, 04:21
spjmanoli wrote: If k is a positive integer, then 20k is divisible by how many different positive integers? (1) k is prime (2) k = 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? Finding the Number of Factors of an IntegerFirst 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. Answer: B. Hope it's clear.
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Re: If k is a positive integer, then 20k is divisible by
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22 Aug 2013, 04:21
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?



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Re: If k is a positive integer, then 20k is divisible by how man
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14 Mar 2016, 00:09



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Re: If k is a positive integer, then 20k is divisible by how man
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25 Jul 2017, 15:41
hellzangl wrote: If k is a positive integer, then 20k is divisible by how many different positive integers? (1) k is prime (2) k = 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? 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. K= 5 x 2 x 2 x (2 x 1) Thus B



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