Which is larger: $1997^{1999}$ or $1999^{1997}$?
# I. Solution.

$$
1997=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\underbrace{1+1+\cdots+1}_{1995 \text{ terms}}
$$

By the arithmetic and geometric mean inequality,

$$
\frac{1997}{1999}=\frac{\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+1+1+\cdots+1}{1999} \geq \sqrt[1999]{\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot 1 \cdots \cdots 1}=\sqrt[1999]{\frac{1}{16}}
$$

Thus,

$$
\left(\frac{1997}{1999}\right)^{1999} \geq \frac{1}{16}
$$

which implies

$$
1997^{1999} \geq 1999^{1999} \cdot \frac{1}{16} \geq 1999^{1997} \cdot \frac{1999^{2}}{16}
$$

Since $\frac{1999^{2}}{16}>1$, it follows that $1997^{1999}>1999^{1997}$.

(Tarcsi Károly, Szeged, Radnóti M. Experimental Gymnasium, 10th grade)

II. Solution. We prove that $(n+2)^{n}<n^{n+2}$ for all natural numbers $n \geq 3$. It suffices to prove the equivalent inequality $\sqrt[n+2]{n+2}<\sqrt[n]{n}$, which means we need to show that the sequence $\{\sqrt[n]{n}\}$ is strictly decreasing for $n \geq 3$. We need to show that $\sqrt[n+1]{n+1}<\sqrt[n]{n}$, or equivalently,

$$
\left(1+\frac{1}{n}\right)^{n}<n
$$

We prove this by induction: for $n=3$, (1) indeed holds, since $\left(1+\frac{1}{3}\right)^{3}=\frac{64}{27}<3$. Now let $n \geq 3$, and assume that (1) holds. We will show that $\left(1+\frac{1}{n+1}\right)^{n+1}<n+1$ also holds. Indeed,

$$
\begin{aligned}
& \left(1+\frac{1}{n+1}\right)^{n+1}=\left(1+\frac{1}{n+1}\right)^{n}\left(1+\frac{1}{n+1}\right)< \\
& <\left(1+\frac{1}{n}\right)^{n}\left(1+\frac{1}{n+1}\right)<\frac{n(n+2)}{n+1}<n+1
\end{aligned}
$$

This completes the proof, so it is indeed true that $(n+2)^{n}<n^{n+2}$, and thus $1999^{1997}<1997^{1999}$.

(Kovács Erika Renáta, Hajdúszoboszló, Hőgyes E. Gymnasium, 8th grade)

III. Solution. Let's calculate how many digits each number has. For any positive integer $x$, the number of digits is $[\lg x+1]$.

$$
\begin{aligned}
& {\left[\lg 1997^{1999}+1\right]=[1999 \cdot \lg 1997+1]=6598} \\
& {\left[\lg 1999^{1997}+1\right]=[1997 \cdot \lg 1999+1]=6592}
\end{aligned}
$$

Thus, $1997^{1999}$ has more digits than $1999^{1997}$, so it is definitely larger.

(Máthé András, Budapest, ELTE Apáczai Csere J. Gymnasium, 10th grade)