How to solve equations involving multiple floors of the same variable?

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I want to solve equations of this type:



$$lfloor1xrfloor+lfloor2xrfloor+lfloor3xrfloor+lfloor4xrfloor+lfloor5xrfloor=10$$




algebra-precalculus




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  • 3




    Good for you. To that end, what have you tried? And did you have a question?
    – amWhy
    2 days ago










  • I do not know where to start at all. Adding them seems terrible, and the only way I can think is brute-forcing it...
    – Vivek Sivaramakrishnan
    2 days ago






  • 4




    Hint: $x-1<lfloor x rfloorleq x$. We can estimate $x$.
    – Rumpelstiltskin
    2 days ago











  • If nothing else you can narrow the range of possible solutions, as @Adam suggests. Note that the left hand side is a (weakly) increasing function of $x$. E.g. since setting $x=10$ gives a value that is too large, any solution must be... less than $10$.
    – hardmath
    2 days ago










  • @hardmath $x = 1$ is already too large.
    – Theo Bendit
    2 days ago














up vote
4
down vote

favorite
2












I want to solve equations of this type:



$$lfloor1xrfloor+lfloor2xrfloor+lfloor3xrfloor+lfloor4xrfloor+lfloor5xrfloor=10$$




algebra-precalculus




share|cite|improve this question

















  • 3




    Good for you. To that end, what have you tried? And did you have a question?
    – amWhy
    2 days ago










  • I do not know where to start at all. Adding them seems terrible, and the only way I can think is brute-forcing it...
    – Vivek Sivaramakrishnan
    2 days ago






  • 4




    Hint: $x-1<lfloor x rfloorleq x$. We can estimate $x$.
    – Rumpelstiltskin
    2 days ago











  • If nothing else you can narrow the range of possible solutions, as @Adam suggests. Note that the left hand side is a (weakly) increasing function of $x$. E.g. since setting $x=10$ gives a value that is too large, any solution must be... less than $10$.
    – hardmath
    2 days ago










  • @hardmath $x = 1$ is already too large.
    – Theo Bendit
    2 days ago












up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
2






2





I want to solve equations of this type:



$$lfloor1xrfloor+lfloor2xrfloor+lfloor3xrfloor+lfloor4xrfloor+lfloor5xrfloor=10$$




algebra-precalculus




share|cite|improve this question













I want to solve equations of this type:



$$lfloor1xrfloor+lfloor2xrfloor+lfloor3xrfloor+lfloor4xrfloor+lfloor5xrfloor=10$$





algebra-precalculus





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago
























asked 2 days ago









Vivek Sivaramakrishnan

234




234







  • 3




    Good for you. To that end, what have you tried? And did you have a question?
    – amWhy
    2 days ago










  • I do not know where to start at all. Adding them seems terrible, and the only way I can think is brute-forcing it...
    – Vivek Sivaramakrishnan
    2 days ago






  • 4




    Hint: $x-1<lfloor x rfloorleq x$. We can estimate $x$.
    – Rumpelstiltskin
    2 days ago











  • If nothing else you can narrow the range of possible solutions, as @Adam suggests. Note that the left hand side is a (weakly) increasing function of $x$. E.g. since setting $x=10$ gives a value that is too large, any solution must be... less than $10$.
    – hardmath
    2 days ago










  • @hardmath $x = 1$ is already too large.
    – Theo Bendit
    2 days ago












  • 3




    Good for you. To that end, what have you tried? And did you have a question?
    – amWhy
    2 days ago










  • I do not know where to start at all. Adding them seems terrible, and the only way I can think is brute-forcing it...
    – Vivek Sivaramakrishnan
    2 days ago






  • 4




    Hint: $x-1<lfloor x rfloorleq x$. We can estimate $x$.
    – Rumpelstiltskin
    2 days ago











  • If nothing else you can narrow the range of possible solutions, as @Adam suggests. Note that the left hand side is a (weakly) increasing function of $x$. E.g. since setting $x=10$ gives a value that is too large, any solution must be... less than $10$.
    – hardmath
    2 days ago










  • @hardmath $x = 1$ is already too large.
    – Theo Bendit
    2 days ago







3




3




Good for you. To that end, what have you tried? And did you have a question?
– amWhy
2 days ago




Good for you. To that end, what have you tried? And did you have a question?
– amWhy
2 days ago












I do not know where to start at all. Adding them seems terrible, and the only way I can think is brute-forcing it...
– Vivek Sivaramakrishnan
2 days ago




I do not know where to start at all. Adding them seems terrible, and the only way I can think is brute-forcing it...
– Vivek Sivaramakrishnan
2 days ago




4




4




Hint: $x-1<lfloor x rfloorleq x$. We can estimate $x$.
– Rumpelstiltskin
2 days ago





Hint: $x-1<lfloor x rfloorleq x$. We can estimate $x$.
– Rumpelstiltskin
2 days ago













If nothing else you can narrow the range of possible solutions, as @Adam suggests. Note that the left hand side is a (weakly) increasing function of $x$. E.g. since setting $x=10$ gives a value that is too large, any solution must be... less than $10$.
– hardmath
2 days ago




If nothing else you can narrow the range of possible solutions, as @Adam suggests. Note that the left hand side is a (weakly) increasing function of $x$. E.g. since setting $x=10$ gives a value that is too large, any solution must be... less than $10$.
– hardmath
2 days ago












@hardmath $x = 1$ is already too large.
– Theo Bendit
2 days ago




@hardmath $x = 1$ is already too large.
– Theo Bendit
2 days ago










2 Answers
2






active

oldest

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up vote
3
down vote



accepted










Think of it like there are five rotating spinners, where the $i^th$ spinner makes $i$ revolutions per minute. All spinners start at the same position at $x=0$ minutes. The first spinner completes a revolution every $60$ seconds, the second every $30$ seconds, etc. The number of revolutions completed at time $x$ by the $i^th$ spinner is $lfloor ixrfloor$, so your problem is to find all of the times where the combined total of revolutions is $10.$



Make a schedule of all the times a revolution is completed (a "tick"):



Spinner 1 ...........................................................x
Spinner 2 .............................x.............................x
Spinner 3 ...................x...................x...................x
Spinner 4 ..............x..............x..............x..............x
Spinner 5 ...........x...........x...........x...........x...........x


Each . is one second, and each x is a tick. We see that the $10^th$ tick occurs when Spinner 5 makes its penultimate tick at $x=0.8$ minutes $=48$ seconds. Therefore, the set of times for which there are $10$ ticks total is $xin [0.8,1)$.






share|cite|improve this answer




























    up vote
    0
    down vote













    You can start from the following.



    $x=[x]+x$ and consider number of cases:



    1. $0leqx<frac15;$


    2. $frac15leqx<frac25;$


    $.$



    $.$



    $.$



    Good luck!



    Also, the left side increases!






    share|cite|improve this answer























    • Wouldn't you want a finer partition than that? The parts in this partition will make $lfloor 4x rfloor$ annoying to keep track of, for example.
      – Theo Bendit
      2 days ago











    • @Theo Bendit It's the standard notation. $[x]$ and $lfloor xrfloor$ they are the same. There is also $lceil xrceil$ of course, but it's another story. If you say about the essence of my hint then it's the hint only. I did not write all cases.
      – Michael Rozenberg
      2 days ago











    • @MichaelRozenberg $ lfloor x rfloor $ to me means floor function where I have seen $ [x] $ as just being used as brackets, equivalent to $ (x) $.
      – Warren Hill
      2 days ago










    • @Warren Hill We always know from the context about which thing we say.
      – Michael Rozenberg
      2 days ago










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    Think of it like there are five rotating spinners, where the $i^th$ spinner makes $i$ revolutions per minute. All spinners start at the same position at $x=0$ minutes. The first spinner completes a revolution every $60$ seconds, the second every $30$ seconds, etc. The number of revolutions completed at time $x$ by the $i^th$ spinner is $lfloor ixrfloor$, so your problem is to find all of the times where the combined total of revolutions is $10.$



    Make a schedule of all the times a revolution is completed (a "tick"):



    Spinner 1 ...........................................................x
    Spinner 2 .............................x.............................x
    Spinner 3 ...................x...................x...................x
    Spinner 4 ..............x..............x..............x..............x
    Spinner 5 ...........x...........x...........x...........x...........x


    Each . is one second, and each x is a tick. We see that the $10^th$ tick occurs when Spinner 5 makes its penultimate tick at $x=0.8$ minutes $=48$ seconds. Therefore, the set of times for which there are $10$ ticks total is $xin [0.8,1)$.






    share|cite|improve this answer

























      up vote
      3
      down vote



      accepted










      Think of it like there are five rotating spinners, where the $i^th$ spinner makes $i$ revolutions per minute. All spinners start at the same position at $x=0$ minutes. The first spinner completes a revolution every $60$ seconds, the second every $30$ seconds, etc. The number of revolutions completed at time $x$ by the $i^th$ spinner is $lfloor ixrfloor$, so your problem is to find all of the times where the combined total of revolutions is $10.$



      Make a schedule of all the times a revolution is completed (a "tick"):



      Spinner 1 ...........................................................x
      Spinner 2 .............................x.............................x
      Spinner 3 ...................x...................x...................x
      Spinner 4 ..............x..............x..............x..............x
      Spinner 5 ...........x...........x...........x...........x...........x


      Each . is one second, and each x is a tick. We see that the $10^th$ tick occurs when Spinner 5 makes its penultimate tick at $x=0.8$ minutes $=48$ seconds. Therefore, the set of times for which there are $10$ ticks total is $xin [0.8,1)$.






      share|cite|improve this answer























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Think of it like there are five rotating spinners, where the $i^th$ spinner makes $i$ revolutions per minute. All spinners start at the same position at $x=0$ minutes. The first spinner completes a revolution every $60$ seconds, the second every $30$ seconds, etc. The number of revolutions completed at time $x$ by the $i^th$ spinner is $lfloor ixrfloor$, so your problem is to find all of the times where the combined total of revolutions is $10.$



        Make a schedule of all the times a revolution is completed (a "tick"):



        Spinner 1 ...........................................................x
        Spinner 2 .............................x.............................x
        Spinner 3 ...................x...................x...................x
        Spinner 4 ..............x..............x..............x..............x
        Spinner 5 ...........x...........x...........x...........x...........x


        Each . is one second, and each x is a tick. We see that the $10^th$ tick occurs when Spinner 5 makes its penultimate tick at $x=0.8$ minutes $=48$ seconds. Therefore, the set of times for which there are $10$ ticks total is $xin [0.8,1)$.






        share|cite|improve this answer













        Think of it like there are five rotating spinners, where the $i^th$ spinner makes $i$ revolutions per minute. All spinners start at the same position at $x=0$ minutes. The first spinner completes a revolution every $60$ seconds, the second every $30$ seconds, etc. The number of revolutions completed at time $x$ by the $i^th$ spinner is $lfloor ixrfloor$, so your problem is to find all of the times where the combined total of revolutions is $10.$



        Make a schedule of all the times a revolution is completed (a "tick"):



        Spinner 1 ...........................................................x
        Spinner 2 .............................x.............................x
        Spinner 3 ...................x...................x...................x
        Spinner 4 ..............x..............x..............x..............x
        Spinner 5 ...........x...........x...........x...........x...........x


        Each . is one second, and each x is a tick. We see that the $10^th$ tick occurs when Spinner 5 makes its penultimate tick at $x=0.8$ minutes $=48$ seconds. Therefore, the set of times for which there are $10$ ticks total is $xin [0.8,1)$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









        Mike Earnest

        14.5k11644




        14.5k11644




















            up vote
            0
            down vote













            You can start from the following.



            $x=[x]+x$ and consider number of cases:



            1. $0leqx<frac15;$


            2. $frac15leqx<frac25;$


            $.$



            $.$



            $.$



            Good luck!



            Also, the left side increases!






            share|cite|improve this answer























            • Wouldn't you want a finer partition than that? The parts in this partition will make $lfloor 4x rfloor$ annoying to keep track of, for example.
              – Theo Bendit
              2 days ago











            • @Theo Bendit It's the standard notation. $[x]$ and $lfloor xrfloor$ they are the same. There is also $lceil xrceil$ of course, but it's another story. If you say about the essence of my hint then it's the hint only. I did not write all cases.
              – Michael Rozenberg
              2 days ago











            • @MichaelRozenberg $ lfloor x rfloor $ to me means floor function where I have seen $ [x] $ as just being used as brackets, equivalent to $ (x) $.
              – Warren Hill
              2 days ago










            • @Warren Hill We always know from the context about which thing we say.
              – Michael Rozenberg
              2 days ago














            up vote
            0
            down vote













            You can start from the following.



            $x=[x]+x$ and consider number of cases:



            1. $0leqx<frac15;$


            2. $frac15leqx<frac25;$


            $.$



            $.$



            $.$



            Good luck!



            Also, the left side increases!






            share|cite|improve this answer























            • Wouldn't you want a finer partition than that? The parts in this partition will make $lfloor 4x rfloor$ annoying to keep track of, for example.
              – Theo Bendit
              2 days ago











            • @Theo Bendit It's the standard notation. $[x]$ and $lfloor xrfloor$ they are the same. There is also $lceil xrceil$ of course, but it's another story. If you say about the essence of my hint then it's the hint only. I did not write all cases.
              – Michael Rozenberg
              2 days ago











            • @MichaelRozenberg $ lfloor x rfloor $ to me means floor function where I have seen $ [x] $ as just being used as brackets, equivalent to $ (x) $.
              – Warren Hill
              2 days ago










            • @Warren Hill We always know from the context about which thing we say.
              – Michael Rozenberg
              2 days ago












            up vote
            0
            down vote










            up vote
            0
            down vote









            You can start from the following.



            $x=[x]+x$ and consider number of cases:



            1. $0leqx<frac15;$


            2. $frac15leqx<frac25;$


            $.$



            $.$



            $.$



            Good luck!



            Also, the left side increases!






            share|cite|improve this answer















            You can start from the following.



            $x=[x]+x$ and consider number of cases:



            1. $0leqx<frac15;$


            2. $frac15leqx<frac25;$


            $.$



            $.$



            $.$



            Good luck!



            Also, the left side increases!







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago


























            answered 2 days ago









            Michael Rozenberg

            86.9k1575178




            86.9k1575178











            • Wouldn't you want a finer partition than that? The parts in this partition will make $lfloor 4x rfloor$ annoying to keep track of, for example.
              – Theo Bendit
              2 days ago











            • @Theo Bendit It's the standard notation. $[x]$ and $lfloor xrfloor$ they are the same. There is also $lceil xrceil$ of course, but it's another story. If you say about the essence of my hint then it's the hint only. I did not write all cases.
              – Michael Rozenberg
              2 days ago











            • @MichaelRozenberg $ lfloor x rfloor $ to me means floor function where I have seen $ [x] $ as just being used as brackets, equivalent to $ (x) $.
              – Warren Hill
              2 days ago










            • @Warren Hill We always know from the context about which thing we say.
              – Michael Rozenberg
              2 days ago
















            • Wouldn't you want a finer partition than that? The parts in this partition will make $lfloor 4x rfloor$ annoying to keep track of, for example.
              – Theo Bendit
              2 days ago











            • @Theo Bendit It's the standard notation. $[x]$ and $lfloor xrfloor$ they are the same. There is also $lceil xrceil$ of course, but it's another story. If you say about the essence of my hint then it's the hint only. I did not write all cases.
              – Michael Rozenberg
              2 days ago











            • @MichaelRozenberg $ lfloor x rfloor $ to me means floor function where I have seen $ [x] $ as just being used as brackets, equivalent to $ (x) $.
              – Warren Hill
              2 days ago










            • @Warren Hill We always know from the context about which thing we say.
              – Michael Rozenberg
              2 days ago















            Wouldn't you want a finer partition than that? The parts in this partition will make $lfloor 4x rfloor$ annoying to keep track of, for example.
            – Theo Bendit
            2 days ago





            Wouldn't you want a finer partition than that? The parts in this partition will make $lfloor 4x rfloor$ annoying to keep track of, for example.
            – Theo Bendit
            2 days ago













            @Theo Bendit It's the standard notation. $[x]$ and $lfloor xrfloor$ they are the same. There is also $lceil xrceil$ of course, but it's another story. If you say about the essence of my hint then it's the hint only. I did not write all cases.
            – Michael Rozenberg
            2 days ago





            @Theo Bendit It's the standard notation. $[x]$ and $lfloor xrfloor$ they are the same. There is also $lceil xrceil$ of course, but it's another story. If you say about the essence of my hint then it's the hint only. I did not write all cases.
            – Michael Rozenberg
            2 days ago













            @MichaelRozenberg $ lfloor x rfloor $ to me means floor function where I have seen $ [x] $ as just being used as brackets, equivalent to $ (x) $.
            – Warren Hill
            2 days ago




            @MichaelRozenberg $ lfloor x rfloor $ to me means floor function where I have seen $ [x] $ as just being used as brackets, equivalent to $ (x) $.
            – Warren Hill
            2 days ago












            @Warren Hill We always know from the context about which thing we say.
            – Michael Rozenberg
            2 days ago




            @Warren Hill We always know from the context about which thing we say.
            – Michael Rozenberg
            2 days ago












             

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