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That way you can minimize the frustration the floor function brings to the table.

er, i guess the interval should really be [n,n+1)

Domain = [0, 100] :S

wait, the function\[x^2+\lfloor x \rfloor x\]is strictly increasing, isnt it?

then it can have at most one solution.

That means it has 0 :'(

strictly increasing on the interval [0,100] anyways.

no, its strictly increasing for sure.

Sorry!!! Wait... a mistake in the question posted here!!

ah ok, that makes more sense.

I'm sorry!!!! :(

less than >.> i meant greater than lol.

\[x\ge 81.25\]

>.<\[x\ge 81\]

So there are 19 solutions, one in each of the intervals [81, 82), [82,83),....., [99,100)

Thanks! I've never thought about this way to look at the problem.. :(

I'll remember it! Thanks again!!

x= 81.25 /{x}
x can go max to 100
so fractional part can be minimum 0.8125 and max 0.999999...

just substitute in new values of {x} to get new values of x
so according to me,,infinite solutions..

ohh wait,,did i make a mistake in logic,,i am getting a feeling,,!??

yep..i understand better now..thank you..