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  1. #1
    Member grgrsanjay's Avatar
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    this featured in my MATHS OLYMPIAD

    the number of real values for x such that{ (x+1)^3} = x^3 where { x} is the fractional part of x

    i.e{2.635}=0.635

    i think the answer is 1 cuz we get the answer right when we substitute 0 and i cannot think any other values that satisfy this equation

    options are:
    (A)0
    (B)1
    (C)4
    (D)6
    Last edited by grgrsanjay; August 29th 2010 at 07:13 AM. Reason: was wrong with concept so corrected my mistake
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  2. #2
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    [QUOTE=grgrsanjay;552428]
    i.e{2.635}=635

    I think that's wrong.

    {x}=x-[x]
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  3. #3
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    Quote Originally Posted by grgrsanjay View Post
    the number of real values for x such that{ (x+1)^3} = x^3 where { x} is the fractional part of x

    i.e{2.635}=635

    i think the answer is 1 cuz we get the answer right when we substitute 0 and i cannot think any other values that satisfy this equation

    options are:
    (A)0
    (B)1
    (C)4
    (D)6
    If x^3 is the fractional part of x then it must lie between 0 and 1, and therefore so must x.

    Next, (x+1)^3 = x^3 + 3x^2 + 3x + 1. If the fractional part of this is x^3 then the remaining part 3x^2+3x+1 must be an integer. So we want to find numbers x between 0 and 1 that satisfy the equation 3x^2+3x-n=0 for some integer n. The solution to the quadratic equation is x = \frac{-3+\sqrt{9+12n}}6 (neglecting the negative square root because that leads to a negative value for x). That gives these solutions:

    \begin{array}{l|l}n&\multicolumn{1}{c}{x}\\\hline0  &0\\ 1&0.2637...\\2&0.4574...\\ 3&0.6180...\\ 4&0.7583...\\ 5&0.8844...\end{array}

    When n = 6 we get x = 1, which is too large. So there are six solutions in all.
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  4. #4
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    I think if you simply plot \{ (x+1)^3 \} and x^3 for 0 \leq x \leq 1, you will find there are 6 roots (i.e., intersections of the graphs of the two functions).

    Maybe someone else will supply a more analytical approach-- but that's the "quick and dirty" way.

    [edit] Beaten to the punch by Opalq, with a nicer solution![/edit]
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  5. #5
    Member grgrsanjay's Avatar
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    could you ....

    is there any other short cut way these methods r too lengthy
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  6. #6
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    Quote Originally Posted by grgrsanjay View Post
    is there any other short cut way these methods r too lengthy
    if x=0, there is a solution, since (0+1)^3=1.0

    If x=1, we have (1+1)^3=2^3=8\Rightarrow\ (x+1)^3-7=x but this is not a solution.

    x must be less than 1, so that x^3<1

    Hence, the remaining solutions will be

    (x+1)^3-6=x

    (x+1)^3-5=x

    (x+1)^3-4=x

    (x+1)^3-3=x

    (x+1)^3-2=x

    (x+1)^3-1=x

    The last one was already mentioned, so there are 6 solutions.
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