## Floor function problem

Forum for the GRE subject test in mathematics.
lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Floor function problem

How many solutions has the equation below in the interval [1, 100] ?

x^2 - [x^2] = {x}^2

here [x] - floor function, {x} - fractional part.

toughluck
Posts: 9
Joined: Thu Jun 25, 2009 10:57 am

### Re: Floor function problem

So I'm not 100% sure on this but I believe that the answer is one plus the sum of the series from n=1 to n=99 of 2n.

I came up with the following inequality (y+b^2)/(2b)<sqrt(y+1) where y is every whole number from 1 to 10000 and b is [sqrt(y)]. Where [] indicate the floor function.

mk
Posts: 13
Joined: Fri May 22, 2009 1:14 am

### Re: Floor function problem

I don't have an exact numerical answer, although I believe I have found a way to a solution.

Since we are working with positive numbers, x^2 - [x^2] = {x^2} (any number is the sum of its whole and fractional parts). Substituting, we now have the equation {x^2} = {x}^2. Substituting x = [x] + {x}, we have {x}^2 = {([x] + {x})^2} = {[x]^2 + 2[x]{x} + {x}^2}. Since [x]^2 is an integer, we can take it out of the inside since it won't affect the fractional part. This gives us {2[x]{x} + {x}^2} = {x}^2. This equation has solutions exactly when 2[x]{x} is an integer.

So each integer N between 1 and 100 has a finite number of fractional parts B<1 s.t. N+B is a solution to the equation. N+B will be a solution when 2NB is an integer. B=0 works for any N, and to find the rest of the solutions find all the divisors of 2N not including 0 and 1. B=1/(any divisor of 2N) will make N+B satisfy the equation.

I'm sure there is a nice way to compute the number of divisors of 2N for any N, however I know little number theory and thus will leave it up to those more suited for the task. I do think the reasoning behind the solution is correct though.

Posts: 9
Joined: Tue Sep 29, 2009 1:07 pm

### Re: Floor function problem

Developing on the reasoning of mk (which I revised and found no errors), we must find all real numbers $$x\in [1,100]$$ such that $$2[x]\{x\}\in\mathbb N$$. This implies that $$\{x\} = \frac{n}{2[x]}$$ for some $$n=0,\dots,2[x]-1$$. This limitation occurs because we must have $$\{x\}<1$$. Given $$[x]$$ between 1 and 99, there are $$2[x]$$ possibilities for $$\{x\}$$, and therefore $$2[x]$$ possibilities for $$x = [x]+\{x\}$$. This gives a total of $$\sum_{n=1}^{99} 2n = 9900$$ possibilities for $$x$$ with $$[x]=1,\dots,99$$. There is only one other case to consider, $$x=100$$, so the equation has 9901 solutions in $$[1,100]$$.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

Correct!