I2.1 If the $50^{\text {th }}$ power of a two-digit number $P$ is a 69-digit number, find the value of $P$.
(Given that $\log 2=0.3010, \log 3=0.4771, \log 11=1.0414$.)

I2.2 The roots of the equation $x^{2}+a x-P+7=0$ are $\alpha$ and $\beta$, whereas the roots of the equation $x^{2}+b x-r=0$ are $-\alpha$ and $-\beta$. If the positive root of the equation $\left(x^{2}+a x-P+7\right)+\left(x^{2}+b x-r\right)=0$ is $Q$, find the value of $Q$.
Reference: 1995 HG5 . . $37^{100} \ldots$ 157-digit number, $37^{15} \ldots n$-digit ....
$$
\begin{array}{l}
P^{50}=y, 10<P \leq 99,10^{68} \leq y<10^{69} \\
P=y^{\frac{1}{50}} ; 10^{68 \div 50}<P<10^{69 \div 50} \\
1.34<\log P<1.38 \\
\log 22=\log 2+\log 11=1.3424 ; \log 24=3 \log 2+\log 3=1.3801 \\
\log 22<\log P<\log 24, P=23 \\
\alpha+\beta=-a, \alpha \beta=-16 ;-\alpha-\beta=-b,(-\alpha)(-\beta)=-r \\
\therefore b=-a, r=16 \\
\left(x^{2}+a x-P+7\right)+\left(x^{2}+b x-r\right)=0 \text { is equivalent to }\left(x^{2}+a x-16\right)+\left(x^{2}-a x-16\right)=0 \\
2 x^{2}-32=0 \\
x=4 \text { or }-4 \\
Q=\text { positive root }=4
\end{array}
$$