High School Can You Solve This Math Challenge Without a Calculator?

[anemone]

Compute the value of $\left\lfloor \dfrac{2005^3}{2003\cdot 2004}-\dfrac{2003^3}{2004\cdot 2005} \right\rfloor$ without the help of a calculator where $\lfloor x \rfloor$ denotes the greatest integer less than $x$.

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[anemone]

Congratulations to the following members for their correct solutions!:)

1. kaliprasad
2. MarkFL
3. mente oscura
4. lfdahl

Solution from kaliprasad:

Spoiler

To keep the arithmetic simple put

2004 = x to get

$\frac{(x+1)^3}{x(x-1)} - \frac{(x-1)^3}{x(x+1)}$
= $\frac{(x+1)^4 – (x-1)^4}{x(x+1)(x-1)}$
= $\frac{2 (4x^3 + 4x)}{(x(x+1)(x-1)}$
=$\frac{8x(x^2+1)}{x(x^2-1)}$
= $\frac{8(x^2 + 1)}{(x^2-1)}$
= $8 + \frac{16}{x^2-1}$

Now the beauty is for any x $\ge$ 5 the 2nd term is between 0 and one so integral part is 8

Hence ans is 8.

Alternate solution from mente oscura:

Spoiler

\dfrac{2005^3}{2003\cdot 2004}-\dfrac{2003^3}{2004\cdot 2005}

= \dfrac{1}{2003\cdot 2004\cdot 2005} \ (2005^4-2003^4)

= \dfrac{1}{2003\cdot 2004\cdot 2005} \ (2005^2+2003^2) \ (2005+2003) \ (2005-2003)

= \dfrac{1}{2003\cdot 2005} \ (2005^2+2003^2) \cdot 2\cdot 2

= 4\cdot (\dfrac{2005}{2003}+\dfrac{2003}{2005})

= 4\cdot (1+\dfrac{2}{2003}+1-\dfrac{2}{2005})=4\cdot (2+\dfrac{2}{2003}-\dfrac{2}{2005})

Such that:

0< \dfrac{2}{2003}-\dfrac{2}{2005} < \dfrac{1}{4}

Therefore:

\left\lfloor \dfrac{2005^3}{2003\cdot 2004}-\dfrac{2003^3}{2004\cdot 2005} \right\rfloor= 8