Suppose that $\mathbf{G}=10^{100}$. ( $\mathbf{G}$ is known as a googol.) How many times does the digit 9 occur in the integer equal to $\mathbf{G}-1009^{2}$ ?
We note first that $1009^{2}=(1000+9)^{2}=1000^{2}+2 \cdot 1000 \cdot 9+9^{2}=1018081$. Therefore,

$$
\begin{aligned}
\mathbf{G}-1009^{2} & =10^{100}-1018081 \\
& =\left(10^{100}-1\right)-(1081081-1) \\
& =\underbrace{999 \cdots 999}_{100 \text { digits equal to } 9}-1081080 \\
& =\underbrace{999 \cdots 999}_{93 \text { digits equal to } 9} 9999999-1081080 \\
& =\underbrace{999 \cdots 999}_{93 \text { digits equal to } 9} 8918919
\end{aligned}
$$

and $\mathbf{G}-1009^{2}$ has 96 digits equal to 9 .

ANSWER: 96