Help!!!

Give a solution please......

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

Case 1. X> 0. Let the floor of X be A.

Then A (A+1)> 2013> A (A). If A> 44, the right side is not satisfied. If A <45, then the left side is not satisfied. No solutions in this case.

Case 2. X <0. (Again, let the floor of X be A)

Then $A^{2}> 2013 >A(A+1)$ .

Note that A=-45 is the only solution. This gives x= $-\frac {2013}{45}$ as the only solution.

Joel Tan - 6 years, 5 months ago

Well dude still I'didn't get 2013 as the asnwer even by using a calculator.

jaiveer shekhawat - 6 years, 5 months ago

You might be mistaken. $\lfloor -\frac{2013}{45} \rfloor = \lfloor -44.7333 \ldots \rfloor = -45$ instead of $-44$ as you might have thought. You can then see that $x \lfloor x \rfloor = 2013$ very easily.

Jake Lai - 6 years, 5 months ago

@Jake Lai – Thanks...got the right nerve

jaiveer shekhawat - 6 years, 5 months ago

thank you

math man - 6 years, 5 months ago

It is not complete, as the case for x <0 was not considered. There is one solution, as shown in my solution above.

Joel Tan - 6 years, 5 months ago

jessen wkwkwkw:D

Alfian Edgar - 6 years, 5 months ago

lol

math man - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$