Which is the positive integer that is a three-digit number in both the decimal and octal number systems, and the sum of its digits is fourteen in both cases?
Let the three-digit number in the decimal system be denoted by: $10^{2} a+10 b+c$, where $0 \leq a, b, c \leq 9$ and $a \neq 0$. In the octal system: $8^{2} u+8 v+w$, where $0 \leq u, v, w \leq 7$ and $u \neq 0$. According to the conditions:

(1) $a+b+c=u+v+w=14$, and

(2) $10^{2} a+10 b+c=8^{2} u+8 v+w$.

From (1), we get $c=14-a-b$ and $w=14-u-v$. Substituting these into (2) and rearranging the equation, we obtain:

$$
9(11 a+b)=7(9 u+v) .
$$

Here, the left side is divisible by 9, so the right side must also be divisible by 9. Since 9 and 7 have no common divisor greater than 1, only $9 \mid 9 u+v$ is possible. This implies that $v=0$, since $v \leq 7$ according to the condition. Then from (1), $u+w=14$, and since $u, w \leq 7$, it follows that $u=w=7$, so the sought three-digit number in the octal system is: 707, which, when converted to the decimal system, is $7 \cdot 8^{2}+7=455$, and the sum of the digits is indeed 14.

Hotzi Bernadette, (Eger, Lenkey J. Ált. Isk., 8. évf.)