# Problem Of the Day-2

Here is Yesterday Problem

Yesterday Winner was Akash Agarwal

You must be knowing that If you win the Problem of the day for 4 times in a week – then I will send you a Really interesting book on any topic you want 🙂

Today’s Problem  :

Find all possible values of x satisfying :

[x]/[x-2] – [x-2]/[x] = (8{x} + 12)/([x-2][x])

(.) = Normal bracket
{.} = Fractional part function
[.] = GIF/Floor function

I am a young student who loves math . I like number theory and inequalities part the most , and preparing for Math Olympiads :)

Posted on Wednesday,May 30, 2012, in Problems and tagged , , , . Bookmark the permalink. 6 Comments.

1. regina

-4

2. Aayush Kumar

Solution:-
after seeing equation [x]/[x-2] – [x-2]/[x] = (8{x} + 12)/([x-2][x])
we got that x is an integer as
so {x}=0
so…
Now equation comes
[x]/[x-2] – [x-2]/[x] = (12)/([x-2][x])
now on solving it we get
x=4

3. The equation can be modified as
[x]^2-[x-2]^2 = 8{x}+12;
Now as LHS is an integer, RHS need to be an integer as well.
Thus, 8{x}=integer.
Now;
Let [x]=y;
Then,
4y-4 = 8{x}+16;
y = 2{x}+4;
As y is an integer the possible values of {x} can be 0 or 0.5;
For {x} = 0; y = [x] = 4; Thus x = 4; and
For {x} = 0.5; y = [x] = 5; Thus x = 5+0.5 = 5.5
Thus there are two possible values of x as 4 and 5.5 .

4. keshushivang

Akash :

The equation can be modified as
[x]^2-[x-2]^2 = 8{x}+12;
Now as LHS is an integer, RHS need to be an integer as well.
Thus, 8{x}=integer.
Now;
Let [x]=y;
Then,
4y-4 = 8{x}+16;
y = 2{x}+4;
As y is an integer the possible values of {x} can be 0 or 0.5;
For {x} = 0; y = [x] = 4; Thus x = 4; and
For {x} = 0.5; y = [x] = 5; Thus x = 5+0.5 = 5.5
Thus there are two possible values of x as 4 and 5.5 .

Is absolutely correct 🙂

Congratulations

You have answered 2 times correct 🙂

5. Aayush Kumar

Hmmm sir… Akash.. U r from IIT Roorkee

• Akash

Yes, I am from IIT Roorkee… 🙂