$\begin{gathered}{\left[\begin{array}{l}\text { Riddles } \\ {[\text { Evenness and Oddness }}\end{array}\right]}\end{gathered}$

Decode the riddle

$$
\begin{array}{r}
\begin{array}{r}
44 H \\
\times H H
\end{array} \\
+\begin{array}{r}
4 H 4 H \\
4 H H
\end{array} \\
\hline H H H H H
\end{array}
$$

All digits marked with the letter E are even (not necessarily equal); all digits marked with the letter $\mathrm{H}$ are odd (also not necessarily equal).
For the convenience of further reasoning, let's replace all even numbers with vowels and all odd numbers with consonants, keeping in mind that different letters can correspond to the same digit. The riddle will take the following form:

$$
\begin{aligned}
\times E B \\
\times \quad C D \\
\hline \text { F FUG } \\
+Y H K \\
\hline \text { YMNPR }
\end{aligned}
$$

Note that the numbers AEB and YHK differ in the second digit, so \( \mathrm{C} > 1 \). If \( \mathrm{C} > 3 \) or \( \mathrm{A} > 2 \), the product AEB \(\cdot\) C will be a four-digit number, so \( \mathrm{C} = 3, \mathrm{~A} = 2 \). Therefore, I is no more than 2, which means it is equal to 2. From this, it follows that \( \mathrm{D} = 9 \) (with a smaller value of \( \mathrm{D} \), the expression AEB \(\cdot\) D will be less than 2100, while IFUG \( > 2100 \)). \( \mathrm{Y} = 8 \) (otherwise, the entire product will not be a five-digit number). \( \mathrm{E} = 8 \) (if \( \mathrm{E} < 8 \), then \( \mathrm{YHK} < 810 \)). As B changes from 1 to 9, the number IFUG will change from 2529 to 2601, so \( \mathrm{F} = 5 \), and \( \mathrm{B} < 9 \). \( \mathrm{H} = 5 \) (as B changes from 1 to 7, the number YHK will change from 843 to 861). \( \mathrm{K} = 5 \) (otherwise, the number \( 85 \mathrm{~K} \) will not be divisible by 3).

## Answer

\( 285 \times 39 = 11115 \)