this featured in my MATHS OLYMPIAD

• Aug 29th 2010, 04:53 AM
grgrsanjay
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
• Aug 29th 2010, 05:02 AM
ghostbuster9
[QUOTE=grgrsanjay;552428]
i.e{2.635}=635

I think that's wrong.

{x}=x-[x]
• Aug 29th 2010, 07:25 AM
Opalg
Quote:

Originally Posted by grgrsanjay
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.
• Aug 29th 2010, 07:40 AM
awkward
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.

 Beaten to the punch by Opalq, with a nicer solution![/edit]
• Aug 29th 2010, 05:34 PM
grgrsanjay
could you ....
is there any other short cut way these methods r too lengthy
• Aug 29th 2010, 06:55 PM
Quote:

Originally Posted by grgrsanjay
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.