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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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New post Updated on: 22 Aug 2013, 03:17
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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?

Originally posted by hellzangl on 22 Aug 2013, 03:14.
Last edited by Bunuel on 22 Aug 2013, 03:17, 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  [#permalink]

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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

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  [#permalink]

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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]

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New post 24 Feb 2016, 10:43
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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Re: If k is a positive integer, then 20k is divisible by how man  [#permalink]

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New post 25 Jul 2017, 14:41
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
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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 29 Oct 2018, 00:57
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?


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)
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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 23 Nov 2018, 17:22
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 &nbs [#permalink] 23 Nov 2018, 17:22
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