I1.1 Let $P$ be the unit digit of $3^{2003} \times 5^{2002} \times 7^{2001}$. Find the value of $P$.
I1.2 If the equation $\left(x^{2}-x-1\right)^{x+P-1}=1$ has $Q$ integral solutions, find the value of $Q$.
I1.3 Let $x, y$ be real numbers and $x y=1$.
If the minimum value of $\frac{1}{x^{4}}+\frac{1}{Q y^{4}}$ is $R$, find the value of $R$.
11.4 Let $x_{R}, x_{R+1}, \ldots, x_{K}(K>R)$ be $K-R+1$ distinct positive integers and $x_{R}+x_{R+1}+\ldots+x_{K}=2003$ If $S$ is the maximum possible value of $K$, find the value of $S$. (Reference: 2004 HI4)
$3^{2003} \times 7^{2001}$ is an odd number, and the unit digit of $5^{2002}$ is $5 ; P=5$
The equation is $\left(x^{2}-x-1\right)^{x+4}=1$
Either $x^{2}-x-1=1$ $\qquad$ (1) or $x+4=0$ $\qquad$ (2) or $\left(x^{2}-x-1=-1\right.$ and $x+4$ is even)
(1): $x=2$ or -1 ; (2): $x=4$; (3): $x=0$ or 1 and $x$ is even $\Rightarrow x=0$ only

Conclusion: $x=-4,-1,0,2$
$$
\begin{array}{l}
Q=4 \\
\frac{1}{x^{4}}+\frac{1}{Q y^{4}}=\frac{1}{x^{4}}+\frac{1}{4 y^{4}} \geq 2 \sqrt{\frac{1}{x^{4}} \cdot \frac{1}{4 y^{4}}}=1=R \text { (A.M. } \geq \text { G.M.) } \\
x_{1}+x_{2}+\ldots+x_{K}=2003
\end{array}
$$

For maximum possible value of $K, x_{1}=1, x_{2}=2, \ldots, x_{K-1}=K-1$
$$
\begin{array}{l}
1+2+\ldots+K-1+x_{K}=2003 \\
\frac{(K-1) K}{2}+x_{K}=2003, x_{K} \geq K \\
2003 \geq \frac{(K-1) K}{2}+K \\
4006 \geq K^{2}+K \\
K^{2}+K-4006 \leq 0 \\
\left(K-\frac{-1-\sqrt{1+4 \times 4006}}{2}\right)\left(K-\frac{-1+\sqrt{1+4 \times 4006}}{2}\right) \leq 0 \\
0 \leq K \leq \frac{-1+\sqrt{1+4 \times 4006}}{2} \\
\frac{-1+\sqrt{1+4 \times 4006}}{2} \approx \frac{-1+\sqrt{4 \times 4006}}{2}=\sqrt{4006}-0.5 \geq \sqrt{3969}-0.5=\sqrt{63^{2}}-0.5=62.5
\end{array}
$$

Maximum possible $K=62=S$
$$
1+2+\ldots+62=1953=2003-50 ; 1+2+\ldots+61+112=2003
$$