21 (1263). Which is greater: $\frac{10^{11}+1}{10^{11}+1}$ or $\frac{10^{11}+1}{10^{12}+1}$ ?
Solution. To simplify the calculations, let's make the substitution $10^{10}=a$. Then we get:

$$
\frac{10^{11}+1}{10^{11}+1}=\frac{a+1}{10 a+1} ; \frac{10^{11}+1}{10^{12}+1}=\frac{10 a+1}{100 a+1}
$$

Comparing these fractions has been reduced to comparing the fractions

$$
\frac{a+1}{10 a+1} \text { and } \frac{10 a+1}{100 a+1}
$$

Let's bring the fractions to a common denominator:

$$
\begin{aligned}
& \frac{a+1}{10 a+1}=\frac{(a+1)(100 a+1)}{(10 a+1)(100 a+1)} \\
& \frac{10 a+1}{100 a+1}=\frac{(10 a+1)(10 a+1)}{(100 a+1)(10 a+1)}
\end{aligned}
$$

Let's compare the numerators of the fractions. We have:

$$
\begin{aligned}
& (a+1)(100 a+1)=100 a^{2}+101 a+1 \\
& (10 a+1)(10 a+1)=100 a^{2}+20 a+1
\end{aligned}
$$

Since $101 a>20 a\left(a=10^{10}\right)$, the numerator of the first fraction is greater than the numerator of the second fraction, and since the denominators are equal, then 48

$$
\frac{a+1}{10 a+1}>\frac{10 a+1}{100 a+1}, \text { i.e. } \frac{10^{111}+1}{10^{11}+1}>\frac{10^{11}+1}{10^{12}+1}
$$

Two other methods for solving this problem are provided in $\$ 4$.

Answer: the first fraction is greater than the second.