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# Help!!!

Note by Math Man
2 years, 6 months ago

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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. · 2 years, 6 months ago

Well dude still I'didn't get 2013 as the asnwer even by using a calculator. · 2 years, 6 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. · 2 years, 6 months ago

Thanks...got the right nerve · 2 years, 6 months ago

44

· 2 years, 6 months ago

thank you · 2 years, 6 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. · 2 years, 6 months ago

jessen wkwkwkw:D · 2 years, 6 months ago