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

Jasmin Pineda

Jasmin Pineda

Answered question

2022-06-27

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 = k
and 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 ?

Answer & Explanation

Carmelo Payne

Carmelo Payne

Beginner2022-06-28Added 25 answers

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 + 1
and
3 x 5 2 = 3 ( 6 b + 1 ) 5 2 = 9 b 1
and so on
If 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

Finley Mckinney

Beginner2022-06-29Added 11 answers

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 e
where 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 + 7
This 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

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