I2.3 If $X=\sqrt{(100)(102)(103)(105)+(Q-3)}$ is an integer and $R$ is the units digit of $X$, find the value of $R$.

I2.4 If $S$ is the sum of the last 3 digits (hundreds, tens, units) of the product of the positive roots of $\sqrt{2012} \cdot x^{\log _{2012} x}=x^{R}$, find the value of $S$.
Let $y=102.5$, then
$$
\begin{aligned}
& (100)(102)(103)(105)+(12-3) \\
= & (y-2.5)(y-0.5)(y+0.5)(y+2.5)+9 \\
= & \left(y^{2}-6.25\right)\left(y^{2}-0.25\right)+9
\end{aligned}
$$
$$
\begin{array}{l}
=y^{4}-6.5 y^{2}+\frac{25}{16}+9=y^{4}-6.5 y^{2}+\frac{169}{16} \\
=\left(y^{2}-\frac{13}{4}\right)^{2}=\left(102.5^{2}-\frac{13}{4}\right)^{2}=\left(\frac{205^{2}}{4}-\frac{13}{4}\right)^{2} \\
X=\frac{42025^{2}-13}{4}=10503
\end{array}
$$
$R=$ the units digit of $X=3$
Method $2 X=\sqrt{(100)(102)(103)(105)+9}=\sqrt{(100)(100+5)(100+2)(100+3)+9}$
$$
=\sqrt{\left(100^{2}+500\right)\left(100^{2}+500+6\right)+9}=\sqrt{\left(100^{2}+500\right)^{2}+6\left(100^{2}+500\right)+9}=\left(100^{2}+500\right)+3
$$
$R=$ the units digit of $X=3$
$$
\begin{array}{l}
\log _{2012}\left(\sqrt{2012} \cdot x^{\log _{2012} x}\right)=\log _{2012} x^{3} \\
\frac{1}{2}+\left(\log _{2012} x\right)^{2}=3 \log _{2012} x
\end{array}
$$

Let $y=\log _{2012} x$, then $2 y^{2}-6 y+1=0$
$$
\begin{array}{l}
y=\log _{2012} x=\frac{3 \pm \sqrt{7}}{2} \\
\Rightarrow x=2012^{\frac{3+\sqrt{7}}{2}} \text { or } 2012^{\frac{3-\sqrt{7}}{2}} \\
=2012^{3} \\
\equiv 12^{3}(\bmod 1000) \\
=1728(\bmod 1000) \\
\end{array}
$$

Product of positive roots $=2012^{\frac{3+\sqrt{7}}{2}} \times 2012^{\frac{3-\sqrt{7}}{2}}$
$S=$ sum of the last 3 digits $=7+2+8=17$