mathematical treasure part1

its the time 4 the answer

\frac{1}{\sqrt{m}}=\frac{2}{\sqrt{m}+\sqrt{m}}<\frac{2}{\sqrt{m}+\sqrt{m-1}}

on rationalizing

2(\sqrt{m}-\sqrt{m-1})

so we get

\frac{1}{\sqrt{2}}<2(\sqrt{2}-{\sqrt{1}})

\frac{1}{\sqrt{3}}<2(\sqrt{3}-{\sqrt{2}})

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\frac{1}{\sqrt{10000}}<2(\sqrt{10000}-\sqrt{9999})

alternate terms r cutting so we r left with

A<2(100-1)=198

now again

\frac{1}{\sqrt{m}}=\frac{2}{\sqrt{m}+\sqrt{m}}>\frac{2}{\sqrt{m}+{\sqrt{m+1}}}

==2(\sqrt{m+1}-\sqrt{m})

A>2(\sqrt{10001}-\sqrt{2})>197

So we get

197<A<198

so [A]=197

rohan already solved one this kind b4

let me find ...

is the answer 100.

good question . no idea

Add question otherwise there is no sense in asking to answer

even me i cant see the ques

meko q ni dikh ra

this one`s a pretty good one. let others try

yes
ny contenders ???
i will wait 4 some time to give the answer

if someone is trying tell me
i will not give the solution

is there a way to *solve* this?

197

just a single line for loop