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 .
Probus
_________________
Probus
~You Just Can't beat the person who never gives up~ Babe Ruth