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If k is a positive integer, then 20k is divisible by how man

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

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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

[Reveal] Spoiler:
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?
[Reveal] Spoiler: OA

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 [#permalink]

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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 [#permalink]

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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 = 7

[Reveal] Spoiler:
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.

Answer: B.

Hope it's clear.
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Re: If k is a positive integer, then 20k is divisible by how man [#permalink]

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New post 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 [#permalink]

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

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New post 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

[Reveal] Spoiler:
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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Re: If k is a positive integer, then 20k is divisible by how man   [#permalink] 25 Jul 2017, 15:41
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