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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] New post 22 Aug 2013, 03:14
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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, 03:17, edited 1 time in total.
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Re: If k is a positive integer, then 20k is divisible by [#permalink] New post 22 Aug 2013, 03:21
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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 [#permalink] New post 22 Aug 2013, 03:21
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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 = 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] 22 Aug 2013, 03:21
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