If $lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$ the find $A=lfloor{x^2-6x}rfloor-380$












1














We know that



$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$$



Know how to find $$A=lfloor{x^2-6x}rfloor-380$$
Actually i dont have any ideas. Do you help me with finding $A$?






calculus functions







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  • You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
    – Jacob
    Nov 26 at 2:37


















1














We know that



$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$$



Know how to find $$A=lfloor{x^2-6x}rfloor-380$$
Actually i dont have any ideas. Do you help me with finding $A$?






calculus functions







share|cite|improve this question






















  • You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
    – Jacob
    Nov 26 at 2:37
















1












1








1







We know that



$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$$



Know how to find $$A=lfloor{x^2-6x}rfloor-380$$
Actually i dont have any ideas. Do you help me with finding $A$?






calculus functions







share|cite|improve this question













We know that



$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$$



Know how to find $$A=lfloor{x^2-6x}rfloor-380$$
Actually i dont have any ideas. Do you help me with finding $A$?






calculus functions




calculus functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 at 19:55









user602338

1567




1567












  • You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
    – Jacob
    Nov 26 at 2:37




















  • You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
    – Jacob
    Nov 26 at 2:37


















You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
– Jacob
Nov 26 at 2:37






You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
– Jacob
Nov 26 at 2:37












3 Answers
3






active

oldest

votes


















1














We know (if i correctly understand the notations):
$$
begin{aligned}
1380&le x^2-x< 1381 ,\
1380&le x^2-11x< 1381 ,\
&qquadtext{so we add and divide by $2$ getting...}\
1380&le x^2-6x< 1381 .
end{aligned}
$$
The above solves the problem.



It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)



But such an $x$ does not exist. So any $A$ / no $A$ works...






share|cite|improve this answer





























    2














    Since



    $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$



    Everything inbetween will be the same
    $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
    Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$






    share|cite|improve this answer





























      0














      $lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        We know (if i correctly understand the notations):
        $$
        begin{aligned}
        1380&le x^2-x< 1381 ,\
        1380&le x^2-11x< 1381 ,\
        &qquadtext{so we add and divide by $2$ getting...}\
        1380&le x^2-6x< 1381 .
        end{aligned}
        $$
        The above solves the problem.



        It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)



        But such an $x$ does not exist. So any $A$ / no $A$ works...






        share|cite|improve this answer


























          1














          We know (if i correctly understand the notations):
          $$
          begin{aligned}
          1380&le x^2-x< 1381 ,\
          1380&le x^2-11x< 1381 ,\
          &qquadtext{so we add and divide by $2$ getting...}\
          1380&le x^2-6x< 1381 .
          end{aligned}
          $$
          The above solves the problem.



          It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)



          But such an $x$ does not exist. So any $A$ / no $A$ works...






          share|cite|improve this answer
























            1












            1








            1






            We know (if i correctly understand the notations):
            $$
            begin{aligned}
            1380&le x^2-x< 1381 ,\
            1380&le x^2-11x< 1381 ,\
            &qquadtext{so we add and divide by $2$ getting...}\
            1380&le x^2-6x< 1381 .
            end{aligned}
            $$
            The above solves the problem.



            It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)



            But such an $x$ does not exist. So any $A$ / no $A$ works...






            share|cite|improve this answer












            We know (if i correctly understand the notations):
            $$
            begin{aligned}
            1380&le x^2-x< 1381 ,\
            1380&le x^2-11x< 1381 ,\
            &qquadtext{so we add and divide by $2$ getting...}\
            1380&le x^2-6x< 1381 .
            end{aligned}
            $$
            The above solves the problem.



            It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)



            But such an $x$ does not exist. So any $A$ / no $A$ works...







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 at 20:07









            dan_fulea

            6,2301312




            6,2301312























                2














                Since



                $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$



                Everything inbetween will be the same
                $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
                Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$






                share|cite|improve this answer


























                  2














                  Since



                  $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$



                  Everything inbetween will be the same
                  $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
                  Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$






                  share|cite|improve this answer
























                    2












                    2








                    2






                    Since



                    $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$



                    Everything inbetween will be the same
                    $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
                    Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$






                    share|cite|improve this answer












                    Since



                    $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$



                    Everything inbetween will be the same
                    $$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
                    Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 25 at 20:06









                    Dr. Mathva

                    929315




                    929315























                        0














                        $lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$






                        share|cite|improve this answer


























                          0














                          $lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$






                          share|cite|improve this answer
























                            0












                            0








                            0






                            $lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$






                            share|cite|improve this answer












                            $lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 25 at 20:06









                            Shubham Johri

                            3,603716




                            3,603716






























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