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# Is the positive integer n a multiple of 20?

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Director
Joined: 18 Feb 2019
Posts: 603
Location: India
GMAT 1: 460 Q42 V13
GPA: 3.6
Is the positive integer n a multiple of 20?  [#permalink]

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18 Feb 2019, 11:01
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9
00:00

Difficulty:

55% (hard)

Question Stats:

51% (01:30) correct 49% (01:42) wrong based on 75 sessions

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Is the positive integer n a multiple of 20?

(1) n^3 is a multiple of 60.
(2) 10 is a factor of square root on n.
Director
Status: Manager
Joined: 27 Oct 2018
Posts: 746
Location: Egypt
GPA: 3.67
WE: Pharmaceuticals (Health Care)
Re: Is the positive integer n a multiple of 20?  [#permalink]

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18 Feb 2019, 13:38
1
1
Question: does $$n = 2^2*5^1*k?$$

Statement 1:
$$n^3 = 60m$$
$$n^3 = 2^2*5^1*3^1*m$$
since n is a positive integer, $$n = 2^1*5^1*3^1*q$$ , so n is for sure a multiple of 30.
We don't know whether q is an even number, so insufficient.

Statement 2:
$$\sqrt{n} = 10p$$
$$n = 100p^2$$ , so n is divisible by 20 regardless of the value of p --> sufficient

B
Manager
Joined: 11 Dec 2013
Posts: 120
Location: India
GMAT Date: 03-15-2015
WE: Education (Education)
Re: Is the positive integer n a multiple of 20?  [#permalink]

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18 Feb 2019, 22:09
Statement 1 :
$$n^3=1000$$
n=10
Or
$$n^3=4000 , n=20$$
Not sufficient
Statement 2 :
$$\sqrt{n}=10k$$
$$n=100k^2$$
Sufficient
B
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3158
Re: Is the positive integer n a multiple of 20?  [#permalink]

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18 Feb 2019, 22:24
1

Solution

Given:
• The number n is a positive integer

To find:
• Whether n is a multiple of 20 or not

Approach and Working:
If we express 20 in terms of its prime factors, we get 20 = 2^2 x 5^1
• Therefore, if n is a multiple of 20, then n must have 2^2 and 5 as its factors.

Analysing Statement 1
As per the information given in statement 1, n^3 is a multiple of 60.
• Or, n^3 = 2^2 x 3 x 5 x k, where k is a positive integer
• From this, we can say that k must have at least 2, 3^2 and 5^2.
• However, we cannot say whether n is a multiple of 20 or not.

Hence, statement 1 is not sufficient to answer the question.

Analysing Statement 2
As per the information given in statement 2, 10 is a factor of square root of n.
• Therefore, we can say that 100 should be a factor of n.
• As 100 is a multiple of 20, we can say that n is also a multiple of 20.

Hence, statement 2 is sufficient to answer the question.

Hence, the correct answer is option B.

_________________
Intern
Joined: 11 May 2019
Posts: 19
Re: Is the positive integer n a multiple of 20?  [#permalink]

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01 Aug 2019, 00:38
1
Statement 1.
N^3/60 = integer (stated)
So least value of n for which it can be divided by 60 is n= 30 which is not divisible by 20 .
And if n=60 which is divisible by 20 .
Therefore 2 values and insufficient.

Statement 2.
Rootn/10 =integer
Squaring n/100 = integer
Factors of 100 are also factors of n therefore sufficient.

Posted from my mobile device
Intern
Joined: 01 Dec 2018
Posts: 43
Re: Is the positive integer n a multiple of 20?  [#permalink]

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01 Aug 2019, 01:06
Mahmoudfawzy83 wrote:
Question: does $$n = 2^2*5^1*k?$$

Statement 1:
$$n^3 = 60m$$
$$n^3 = 2^2*5^1*3^1*m$$
since n is a positive integer, $$n = 2^1*5^1*3^1*q$$ , so n is for sure a multiple of 30.
We don't know whether q is an even number, so insufficient.

Statement 2:
$$\sqrt{n} = 10p$$
$$n = 100p^2$$ , so n is divisible by 20 regardless of the value of p --> sufficient

B

Hi! Can you please explain that how why did you take all the prime factors power as 1 ?
Director
Status: Manager
Joined: 27 Oct 2018
Posts: 746
Location: Egypt
GPA: 3.67
WE: Pharmaceuticals (Health Care)
Is the positive integer n a multiple of 20?  [#permalink]

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01 Aug 2019, 01:29
2
kunalbean wrote:
Hi! Can you please explain that how why did you take all the prime factors power as 1 ?

lets try a simplified one first,

if $$n^3$$ is a multiple of 2,
then $$n^3 = 2a$$ , where a is an integer.

remember that n is also an integer, and $$n = \sqrt[3]{2a}$$
this constrain forces us to conclude that $$a = 2^2b$$ , where $$b$$ is an integer.
why we concluded that? because otherwise $$\sqrt[3]{2}$$ will produce an irrational number.

back to our question,
$$n^3 = 2^2*5^1*3^1*m$$
so m must be equal to $$2^15^23^2p$$, where p is an integer
and n must have 2,5,3 as primes of it
Is the positive integer n a multiple of 20?   [#permalink] 01 Aug 2019, 01:29
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