33. Let $L$ denote the minimum value of the quotient of a 3-digit number formed by three distinct digits divided by the sum of its digits. Determine $\lfloor 10 L\rfloor$.
33. Answer: 105
A three-digit number can be expressed as $100 a+10 b+c$, and so we are minimizing
$$
F(a b c)=\frac{100 a+10 b+c}{a+b+c}
$$

Observe that with distinct digits $a b c, F(a b c)$ has the minimum value when $a<$ $b<c$. Thus we assume that $0<a<b<c \leq 9$.
Note that
$$
F(a b c)=\frac{100 a+10 b+c}{a+b+c}=1+\frac{99 a+9 b}{a+b+c}
$$

We observe now that $F(a b c)$ is minimum when $c=9$.
$$
F(a b 9)=1+\frac{99 a+9 b}{a+b+9}=1+\frac{9(a+b+9)+90 a-81}{a+b+9}=10+\frac{9(10 a-9)}{a+b+9}
$$

Now $F(a b 9)$ is minimum when $b=8$.
$$
F(a 89)=10+\frac{9(10 a-9)}{a+17}=10+\frac{90(a+17)-1611}{a+17}=100-\frac{1611}{a+17}
$$
which has the minimum value when $a=1$, and so $L=F(189)=105$ and $10 L=105$.