Solving equations including floor function.I got a little trouble solving equations that involve floor function in an efficient way.For example : ⌊ x + 3 2 ⌋ = 4 x + 5 3 In the one above, I get that you basically let 4 x + 5 3 = kand then inserting k in the left side, take k = 8 l , 8 l + 1 and so on and test it.If there's a better solution to the one above plese tell me.My main problem is when it comes down to functions that have multiple floors such as : ⌊ x + 1 3 ⌋ + ⌊ 2 x + 5 6 ⌋ = 3 x − 5 2 Using the same method for each of them and then intersecting the solutions should give me the right answer but is there a faster way to solve equations like this ?

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Solving equations including floor function.I got a little trouble solving equations that involve floor function in an efficient way.For example :      ⌊                  x        +        3            2        ⌋    =            4      x      +      5        3  In the one above, I get that you basically let            4      x      +      5        3    =  kand then inserting k in the left side, take   k  =  8  l  ,  8  l  +  1 and so on and test it.If there&#039;s a better solution to the one above plese tell me.My main problem is when it comes down to functions that have multiple floors such as :      ⌊                  x        +        1            3        ⌋    +      ⌊                  2        x        +        5            6        ⌋    =            3      x      −      5        2  Using the same method for each of them and then intersecting the solutions should give me the right answer but is there a faster way to solve equations like this ?

Carmelo Payne · Accepted Answer

As the left hand side is integer, so should be                     3        x        −        5            2          ⟺    2      |    3  (  x  −  1  )    ⟺    2      |    (  x  −  1  )    ⟹    x is odd (assuming x to be an integer)Again as lcm   (  3  ,  6  ) we need to test for   x  ≡  0  ,  1  ,  2  ,  3  ,  4  ,  5    (  mod    6  ) But as x is odd,   x  ≡  1  ,  3  ,  5    (  mod    6  ) If   x  =  6  b  +  1       ⌊                  x        +        1            3        ⌋    +      ⌊                  2        x        +        5            6        ⌋    =      ⌊                  6        b        +        1        +        1            3        ⌋    +      ⌊                  2        (        6        b        +        1        )        +        5            6        ⌋    =  2  b  +  (  2  b  +  1  )  =  4  b  +  1and                    3        x        −        5            2        =                    3        (        6        b        +        1        )        −        5            2        =  9  b  −  1and so onIf x is not necessarily an integer,                     3        x        −        5            2        +  I    ⟺    x  =                    5        +        2        I            3      Check for   I  ≡  0  ,  1  ,  2    (  mod    3  )

Finley Mckinney · Accepted Answer

Since                     4        x        +        5            3        =      ⌊                            x          +          3                2              ⌋   is an integer, we need 3 to divide   4  x  +  5, i.e.,                    4        x        +        5            3        =  m  ∈      Z      ⟹    x  =                    3        m        −        5            4         where   m  ∈      Z  Hence,      ⌊                            x          +          3                2              ⌋    =  m    ⟹                      x        +        3            2        =  m  +  e    ⟹    x  +  3  =  2  m  +  2  e    ⟹    x  =  2  m  −  3  +  2  ewhere   e  ∈  [  0  ,  1  ). Hence, we need                    3        m        −        5            4        =  2  m  −  3  +  2  e    ⟹    3  m  −  5  =  8  m  −  12  +  8  e    ⟹    8  e  =  −  5  m  +  7This gives us   e  =                    7        −        5        m            8      . Since   e  ∈  [  0  ,  1  ) , we have   7  −  5  m  ∈  [  0  ,  8  )    ⟹    m  =  0  ,  1. Hence,   x  =  −            5      4        ,  −            1      2

Solving equations including floor function. I got a little trouble solving equations that involve f

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