35. Let $x$ be a positive integer, and write $a=\left\lfloor\log _{10} x\right\rfloor$ and $b=\left\lfloor\log _{10} \frac{100}{x}\right\rfloor$. Here $\lfloor c\rfloor$ denotes the greatest integer less than or equal to $c$. Find the largest possible value of $2 a^{2}-3 b^{2}$.
35. Answer: 24
Write $u=\log _{10} x$. Then $\log _{10} \frac{100}{x}=2-u$. Since $a=\left\lfloor\log _{10} x\right\rfloor$, we have
$$
u=a+\gamma \text { for some } 0 \leq \gamma<1 \text {. }
$$

Similarly, since $b=\lfloor 2-u\rfloor$, we have
$$
2-u=b+\delta \text { for some } 0 \leq \delta<1 \text {. }
$$

Then $0 \leq \gamma+\delta<2$. Since $\gamma+\delta=u-a+(2-u-b)=2-a-b$ is an integer, it follows that $\gamma+\delta=0$ or $\gamma+\delta=1$.
Case $1: \gamma+\delta=0$.
Then $\gamma=0$ and $\delta=0$, since $\gamma \geq 0$ and $\delta \geq 0$. Therefore
$$
\begin{aligned}
2 a^{2}-3 b^{2} & =2 u^{2}-3(2-u)^{2} \\
& =-u^{2}+12 u-12 \\
& =24-(u-6)^{2} \leq 24,
\end{aligned}
$$
and the maximum value is attained when $u=6$.
Case $2: \gamma+\delta=1$.
Then we must have $0<\gamma, \delta<1$ by (1) and (2). Also, by (1) and (2), we have $b=\lfloor 2-u\rfloor=\lfloor 2-a-\gamma\rfloor=1-a$. Thus
$$
\begin{aligned}
2 a^{2}-3 b^{2} & =2 a^{2}-3(1-a)^{2} \\
& =-a^{2}+6 a-3 \\
& =6-(a-3)^{2} \leq 6 .
\end{aligned}
$$

Hence the largest possible value of $2 a^{2}-3 b^{2}$ is 24 , when $x=10^{6}$.