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

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About shivang1729

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. 4 is the answer….
    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

  2. 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 .

  3. 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 🙂

  4. Hmmm sir… Akash.. U r from IIT Roorkee

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