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Re: If k is a positive integer, then 20k is divisible by [#permalink]

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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. _________________

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 how man [#permalink]

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27 Oct 2014, 08:18

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Re: If k is a positive integer, then 20k is divisible by how man [#permalink]

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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 [#permalink]

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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

gmatclubot

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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