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The positive integer n is divisible by 25. If n^1/2 is

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Director
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Re: The positive integer n is divisible by 25. If n^1/2 is  [#permalink]

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New post 22 Jun 2017, 06:16
Thank you Bunuel for posting such helpful links on inequalities.
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Re: The positive integer n is divisible by 25. If n^1/2 is  [#permalink]

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New post 22 Jun 2017, 06:24
Shiv2016 wrote:
Thank you Bunuel for posting such helpful links on inequalities.


Bunuel 's questions are really good and very near to GMAT. Solving the questions really help to boost my confidence.
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Re: The positive integer n is divisible by 25. If n^1/2 is  [#permalink]

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New post 21 Feb 2019, 12:10
Hi,

Here are my two cents for this question.
All other members have explained it superbly but if some one is not comfortable with in equations we can view this question as below

We are given that
\(\frac {n}{25}\)= K , where K is integer.

n= \(5^{2}\) * K

where K = \(A^{m}\)* \(B^{n}\), where where A and B are prime numbers , and m and n are positive powers of prime numbers.

Which means when n is written in the form of prime factors it can be written as n= \(5^{2+a}\) * \(A^{m}\)* \(B^{n}\)* .....


We are also given that \(\sqrt{n}\) > 25
which means n > 625
n > \(5^{4}\)

So rewriting n again as product of prime factors we can say n =\(5^{2+b}\) * \(A^{m}\)* \(B^{n}\)*....., where b can take any value from \(0\leq{b}\) < \(\infty\) , but at least either m or n need to take values from \(1\leq{m}\) < \(\infty\) or \(1\leq{n}\) < \(\infty\)
where product of \(A^{m}\)* \(B^{n}\)> 25

We need to know the possible value of \(\frac {n}{25}\)

So we have

\(\frac {n}{25}\) = \(\frac { 5^{2+b} * A^{m}* B^{n}*.....}{25}\)

=\({ 5^{b} * A^{m}* B^{n}*.....}\)

which is clearly more than 25

The only option that is more than 25 is 26

Hence Option E .


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Re: The positive integer n is divisible by 25. If n^1/2 is   [#permalink] 21 Feb 2019, 12:10

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