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











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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 $A$?










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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

















up vote
1
down vote

favorite












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 $A$?










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















up vote
1
down vote

favorite









up vote
1
down vote

favorite











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 $A$?










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






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asked Nov 25 at 19:55









user602338

1326




1326












  • 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

















up vote
1
down vote



accepted










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




























    up vote
    2
    down vote













    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




























      up vote
      0
      down vote













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






      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








        up vote
        1
        down vote



        accepted










        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

























          up vote
          1
          down vote



          accepted










          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























            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            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

            6,1451312




            6,1451312






















                up vote
                2
                down vote













                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

























                  up vote
                  2
                  down vote













                  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























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    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

                    742114




                    742114






















                        up vote
                        0
                        down vote













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






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













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






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









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






                            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

                            1,28239




                            1,28239






























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