Task 2 - 250612

$$
\begin{array}{cccccc} 
& \mathrm{m} & \mathrm{a} & \mathrm{t} & \mathrm{h} & \mathrm{e} \\
+ & & \mathrm{o} & \mathrm{l} & \mathrm{y} & \mathrm{m} \\
+ & & & & \mathrm{p} & \mathrm{i} \\
+ & & & \mathrm{a} & \mathrm{d} & \mathrm{e} \\
\hline \mathrm{k} & \mathrm{l} & \mathrm{a} & \mathrm{s} & \mathrm{s} & \mathrm{e}
\end{array}
$$

In the cryptogram shown, digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) are to be entered for the letters such that the same digits stand for the same letters and the calculation is correct. Furthermore, the following is required:

(1) It holds that $o=m$ and $p=t$ and $y=a$, while for different letters, different digits must be used otherwise.

(2) $a$ is two-thirds of $m$.

(3) $e$ is two-thirds of $a$.

(4) The sum of $a$ and $s$ is equal to $m$.

(5) $d$ is less than $h$.

a) Show that there is exactly one entry that satisfies all these requirements, and give this entry!

b) How many such entries are there if requirement (5) is omitted?
a) If an entry satisfies all the requirements of the task, the following follows:

Due to (2) and (3), both $m$ and $a$ are divisible by 3, with $a$ being two-thirds of $m$. Consequently, $m$ is even divisible by 9.

Since $m$ cannot be 0 as the leading digit, it follows that $m=9$.

From (2) and (3), it follows that $a=6$ and $e=4$. From (4), it then follows that $6+s=9$, so $s=3$. Clearly, $k=1$ and $l=0$ (since the sum is less than $96994+9969+99+694=107756$). Considering (1), we get:

|  | 9 | 6 | $\mathrm{t}$ | $\mathrm{h}$ | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| + |  | 9 | 0 | 6 | 9 |
| + |  |  |  | $\mathrm{t}$ | $\mathrm{i}$ |
| + |  |  | 6 | $\mathrm{~d}$ | 4 |
| 1 | 0 | 6 | 3 | 3 | 4 |

Only for $i=7$ does the sum of the units digits end in 4, with a carry of 2.

For $h, t$, and $d$, the digits 2, 5, and 8 remain, whose sum is 15, so the sum of the tens digits plus the carry of 2 totals 23. Considering the new carry from 23, it follows from the sum of the hundreds digits that $t=5$. This leaves either $d=2$ and $h=8$ or $d=8$ and $h=2 .\left({ }^{*}\right)$

Due to (5), the latter case is excluded. Therefore, only the entry

$$
\begin{array}{rrrrrr} 
& 9 & 6 & 5 & 2 & 4 \\
+ & & 9 & 0 & 6 & 9 \\
+ & & & & 5 & 7 \\
+ & & & 6 & 8 & 4 \\
\hline 1 & 0 & 6 & 3 & 3 & 4
\end{array}
$$

can satisfy all the requirements of the task. Since all requirements are indeed confirmed for this entry, it is proven that there is exactly one such entry of the required type.

b) If we waive the requirement (5), then both cases mentioned in $\left(^{*}\right)$ can occur, and there are exactly two solutions to the cryptogram.