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Give a solution please......

Note by Math Man 2 years, 9 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.

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

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 – Thanks...got the right nerve

44

thank you

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

jessen wkwkwkw:D

lol

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TopNewestCase 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.

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

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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.

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44

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thank you

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

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jessen wkwkwkw:D

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lol

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