Sokan látták a televízió Forsyte Saga filmsorozatát és tudják, hogy Jolyon bácsinak négy unokája volt. Ebből készült az alábbi feladat. - Írjunk az oldalt álló séma betűi helyére számjegyeket - azonos betúk helyére egyenlóket, különbözók helyére különbözőket - úgy, hogy az összeadás helyes legyen.

|  | J |  |
| :---: | :---: | :---: |
| J | 0 | [ |
| $\mathrm{H}$ | O | L |
|  |  | ) |

Many have seen the TV series "The Forsyte Saga" and know that Uncle Jolyon had four grandchildren. From this, the following task was created. - Write digits in place of the letters in the vertical schema - the same digits for the same letters, different digits for different letters - so that the addition is correct.

|  | J |  |
| :---: | :---: | :---: |
| J | 0 | [ |
| $\mathrm{H}$ | O | L |
|  |  | ) |
The numbers "JOLLY" and "HOLLY" - as five-digit numbers - are less than $10^{5}$, similarly "JUNE" is less than 102000, since $L \geqq 2$. Thus, $U, L<9$, and

$$
s<1900+10900+90900+110=103810
$$

therefore, the digit of $10^{3}$ in $s$ is $2 \leqq L \leqq 3$, and repeating based on the upper limit of $e$

$$
s<1900+10400+90400+110=102810
$$

From this, $L=2$ and

$$
s<1900+10230+90230+110=102470
$$

thus, the digit of $10^{2}$ in $s$ is $3 \leqq Y \leqq 4$.

We will show that based on the above, the "carry" from the last column to the tens column can only be 1. On the one hand, this carry - excluding $N$ from the sum and from the terms - is a round ten and $E+2 Y \geqq 10$. On the other hand, $3 \leqq E \leqq 8$ and $Y \leqq 4$, so $E+2 Y \leqq 16$. Therefore, $E+2 Y=10$, from which $Y=3, E=4$, because the assumption $Y=4$ leads to $E=L$.

In the tens place, similarly, $1+N+2+2$ can only give a carry of 1, from which $N=5$, and finally from the hundreds place $1+U+2+2+1=10+Y$ gives $U=7$, and with the new carry, the thousands place is also correct. As a result, the filling of the given schema is unique:

$$
1754+10223+90223+105=102305
$$

Remarks. 1. We can make our estimate with fewer digits if we cut off the rightmost 4 columns with a line. Then (writing $Q$ instead of $O$ to avoid misunderstandings) $Q L L Y$ and $J U N E$ are less than $10^{4}$, and indeed one of them is less than $9 \cdot 10^{3}$, furthermore $J Q N<10^{3}$, so the sum of the numbers to the right of the line is less than $3 \cdot 10^{4}$, thus the carry $m_{4}$ from the 4th column to the 5th is at most 2. Therefore, to the left of the line

$$
\overline{J Q} \leqq J+H+2 \leqq 9+8+2=17
$$

so $J=1$, and based on this $\overline{J Q} \leqq 1+9+2=12, Q \leqq 2$.

Thus, however, the sum to the right of the line is less than $(2+3+3+1)$ thousand, so $m_{4}=0$, hence $J+H \leqq 1+9=10$; on the other hand, $\overline{J Q} \geqq 10$, so $Q=0$. Etc.

2. Based on the result $J<2$, we uniquely took $J=1$, on which the subsequent steps were based. However, solutions starting with $J=0$ have also been submitted:

$$
\begin{aligned}
& 0976+05112+45112+057=051257 \\
& 0124+06338+56338+062=063862
\end{aligned}
$$

However, allowing this raises questions: What is the meaning of the zero written at the beginning of the numbers? Why is there no zero at the beginning of the addend "HOLLY"? And if there were, why not "JJJ JON" as the fourth addend? Etc.