MHB Can You Solve This Math Challenge Without a Calculator?

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The math challenge involves calculating the expression $\left\lfloor \dfrac{2005^3}{2003\cdot 2004}-\dfrac{2003^3}{2004\cdot 2005} \right\rfloor$ without a calculator. Participants discussed various methods to simplify the expression, ultimately leading to the correct integer result. Several members successfully solved the problem, showcasing different approaches. The solutions highlighted the importance of algebraic manipulation and understanding of floor functions. The challenge emphasizes problem-solving skills in mathematics without reliance on technology.
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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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Congratulations to the following members for their correct solutions!:)

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

Solution from kaliprasad:
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:
\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
 
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