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Is the positive integer n a multiple of 20?

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Is the positive integer n a multiple of 20?  [#permalink]

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New post 18 Feb 2019, 11:01
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Question Stats:

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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.
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Re: Is the positive integer n a multiple of 20?  [#permalink]

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New post 18 Feb 2019, 13:38
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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
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Re: Is the positive integer n a multiple of 20?  [#permalink]

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New post 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
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Re: Is the positive integer n a multiple of 20?  [#permalink]

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New post 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.

Answer: B

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Re: Is the positive integer n a multiple of 20?  [#permalink]

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New post 01 Aug 2019, 00:38
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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.
B. Is the answer

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Re: Is the positive integer n a multiple of 20?  [#permalink]

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New post 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 ?
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Is the positive integer n a multiple of 20?  [#permalink]

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New post 01 Aug 2019, 01:29
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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
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Is the positive integer n a multiple of 20?   [#permalink] 01 Aug 2019, 01:29
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