Let $E$ (a natural number) have all digits as 1, the number of its digits is $2k$, and let $F$ have all digits as 2, the number of its digits is $k$. Calculate the square root of $E-F$.

 

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I. solution. Denoting the number written with $k$ ones by $A$, we get $E$ by writing $A$ again after $A$. Due to the second writing, the place value of each digit in the first writing becomes $10^{k}$ times larger, so $E=A \cdot 10^{k}+A$. On the other hand, $F=2 A$, so $E-F=A\left(10^{k}+1-2\right)=A\left(10^{k}-1\right)$.

Writing the digits in $10^{k}$, a 1 is followed by $k$ zeros, so the number one less than this has all digits as 9, with the number of digits being $k$, thus

$$
E-F=A\left(10^{k}-1\right)=A \cdot 9 A=9 A^{2}
$$

therefore, the sought square root is $3 A$, which is the number written with $k$ threes.

Varga Gabriella (Szombathely, Savaria g. III. o. t.)

II. solution. For $k=1,2,3$, $E-F$ is the square of 3, 33, and 333, respectively. We prove by induction that for every positive integer $k$, the square root of the difference between the numbers $E=E_{k}$ and $F=F_{k}$ mentioned in the problem is the number $G_{k}$ written with $k$ threes. $G_{k}$ is obviously one third of the number written with $k$ nines, which is one less than $10^{k}$, so we want to prove the following conjecture:

$$
\sqrt{E_{k}-F_{k}}=G_{k}=\frac{10^{k}-1}{3}
$$

Assuming that (1) is true for some positive integer $n$. We get $E_{n+1}$ from $E_{n}$ by writing two ones after it, thereby increasing the place value of the previous digits by 100: $E_{n+1}=100 \cdot E_{n}+11$, and similarly $F_{n+1}=10 \cdot F_{n}+2$. Since $F_{n}$ is two thirds of $G_{n}$, using (1) we have

$$
\begin{aligned}
E_{n+1} & -F_{n+1}=100 E_{n}-10 F_{n}+9=100\left(G_{n}^{2}+F_{n}\right)-10 F_{n}+9= \\
& =100 G_{n}^{2}+90 F_{n}+9=100 G_{n}^{2}+60 G_{n}+9=\left(10 G_{n}+3\right)^{2}=G_{n+1}^{2}
\end{aligned}
$$

thus (1) is indeed true for the next integer $n+1$ as well.

Szentgáli Ádám (Budapest, Ady E. g. III. o. t.)