Arrange the following fractions in order of magnitude:

$$
t_{1}=\frac{100^{100}+1}{100^{90}+1}, \quad t_{2}=\frac{100^{99}+1}{100^{89}+1}, \quad t_{3}=\frac{100^{101}+1}{100^{91}+1}, t_{4}=\frac{101^{101}+1}{101^{91}+1}, \quad t_{5}=\frac{101^{100}+1}{101^{90}+1}, \quad t_{6}=\frac{99^{99}+1}{99^{89}+1}, \quad t_{7}=\frac{99^{100}+1}{99^{90}+1}
$$
Each fraction can be easily written as the difference between an integer and a positive real fraction, that is, it can be enclosed between two consecutive integers:

$$
\begin{aligned}
& t_{1}=\frac{100^{100}+100^{10}-\left(100^{10}-1\right)}{100^{90}+1}=100^{10}-\frac{100^{10}-1}{100^{90}+1}>100^{10}-1 \\
& t_{2}=100^{10}-\frac{100^{10}-1}{100^{89}+1}, \quad t_{3}=100^{10}-\frac{100^{10}-1}{100^{91}+1} \\
& t_{4}=101^{10}-\frac{101^{10}-1}{101^{91}+1}, \quad t_{5}=101^{10}-\frac{101^{10}-1}{101^{90}+1} \\
& t_{6}=99^{10}-\frac{99^{10}-1}{99^{89}+1}, \quad t_{7}=99^{10}-\frac{99^{10}-1}{99^{90}+1}
\end{aligned}
$$

(Indeed, for example, $0<100^{10}-1<100^{90}+1$). In the transformation of the first three fractions, the minuend is the same number, as is the numerator of the subtrahend, so these fractions are in increasing order if their subtrahends are in decreasing order, which is true if their denominators are in increasing order, thus $t_{2}<t_{1}<t_{3}$. Similarly, $t_{5}<t_{4}$ and $t_{6}<t_{7}$.

On the other hand, among the minuends, it is clearly true that $99^{10}<100^{10}-1<100^{10}<101^{10}-1<101^{10}$. Therefore, the increasing order of the fractions is:

$$
t_{6}, t_{7}, t_{2}, t_{1}, t_{3}, t_{5}, t_{4}
$$

Katalin Péli (Makó, József A. G. II. o. t.)