A three-digit number's digits sum up to 20. If we subtract 16 from the number, half of the resulting difference is a three-digit number whose digits are in reverse order of the original number's digits. Which number is this?
I. solution: Denoting the digits of the sought number by $x, y, z$, according to the problem,

$$
\begin{gathered}
\frac{100 x+10 y+z-16}{2}=100 z+10 y+x \\
x+y+z=20
\end{gathered}
$$

Rearranging (1),

$$
98 x-10 y-199 z=16
$$

Adding 10 times (2) to $\left(1^{\prime}\right)$,

$$
108 x-189 z=216
$$

Simplifying by 27,

$$
4 x-7 z=8
$$

It is immediately clear that $z$ must be divisible by 4. Thus, $z=4 u$, and then $x=\frac{8+7 z}{4}=\frac{8+28 u}{4}=2+7 u$, and $y=20-x-z=20-2-7 u-4 u=18-11 u$.

Since $x, y, z$ are positive single-digit numbers, only $u=1$ fits, and thus

$$
x=2+7 u=9 ; \quad y=18-11 u=7 ; \quad z=4 u=4
$$

Therefore, the sought number is 974.

Morelli Edit (Bp. III., Korvin Ottó-téri lg. I. o. t.)

II. solution: Let the three-digit number be $\overline{a b c}$. According to the problem,

$$
\begin{aligned}
& \frac{\overline{a b c}-16}{2}=\overline{c b a} \\
& a+b+c=20
\end{aligned}
$$

From (3), it follows that either a) $\frac{a}{2}=c$, or b) $\frac{a-1}{2}=c$. From (4), $a+c=20-b$, and thus, since $b \leq 9$,

$$
a+c \geq 11
$$

a) If $a=2 c \leq 9$, then by (5) $3 c \geq 11$, so $c=4$. Then $a=2 \cdot 4=8$, and $b=20-(4+8)=8$. However, 884 does not satisfy our conditions.
b) If $a=2 c+1 \leq 9$, then by (5) $3 c+1 \geq 11$. Again, only $c=4$ fits. $a=2 \cdot 4+1=9, b=20-(4+9)=7$.

Indeed, 974 satisfies the requirements.

Újvári-Menyhárt Zoltán (Bp., 3. sz. Magasép. ip. techn. I. o. t.)