121. From all the digits from 1 to 9, form three three-digit numbers whose product is

a) the smallest; b) the largest.
121. a) The first digits of the three sought numbers must be the smallest; therefore, these numbers in decimal notation have the form:

$$
\overrightarrow{1 A a}, \overrightarrow{2 B b}, \overrightarrow{3 C c},
$$

where the symbol $\overline{x y z}$ denotes the number written with the digits $x, y, z$.

We will prove that: 1) $A < B$, otherwise we would have $\overline{A a} \geq \overline{B b}$ and $\overline{1 A a} \cdot \overline{2 B b}-\overline{2 A a} \cdot \overline{1 B b}=(100+\overline{A a}) \cdot(200+\overline{B b})-(200+\overline{A a}) \times$ $\times(100+\overline{B b})=100(\overline{A a}-\overline{B b})>0$ and, consequently, $\overline{1 B b} \cdot \overline{2 A t} \times$ $\times \overline{3 C c}b$, then we would have

$\overline{1 A a} \cdot \overline{2 B b}-\overline{1 A b} \cdot \overline{2 B a}=(10 \cdot \overline{1 A}+a)(10 \cdot \overline{2 B}+b)-$

![](https://cdn.mathpix.com/cropped/2024_05_21_b937612abb2773f7abffg-160.jpg?height=45&width=892&top_left_y=1318&top_left_x=147)

and, consequently, $\overline{1 A b} \cdot \overline{2 B} a \cdot \overline{3^{C} C}a, C=a+x$, where $x>0$ (due to 1) and 2) $C$ is the largest of the digits $A, B, C$, and $a$ is the smallest of the digits $a, b, c$), then we would have

$$
\begin{aligned}
\overline{1 A a} \cdot \overrightarrow{3 C c}-\overline{1 A c} \cdot \overrightarrow{3 a c}=\sqrt{A a} \cdot(\overline{3 a c}+10 x)- & (\overline{1 A a}+x)(\overline{3 a c})= \\
. & =x(10 \cdot \overline{1 A a}-\overline{3 a c})>0
\end{aligned}
$$

and, consequently, $\overline{1 A C} \cdot \overline{2 B b} \cdot \overline{3 a c}<\overline{1 A a} \cdot \overline{2 B b} \cdot \overrightarrow{3 C c}$.

From 1), 2), and 3) it follows that

$$
A<B<C<a<b<c
$$

## and the sought product has the form

$$
147 \cdot 258 \cdot 369
$$

b) The sought product must have the form

$$
\overline{9 A a}^{-} \cdot \overline{8 B b} \cdot \overline{7 C c}
$$

Analogously to the solution of part a), we can prove that 1) $A < B < C$, 2) $a < b < c$, 3) each of the digits $a, b, c$ is smaller than each of the digits $A, B, C$.

From 1), 2), and 3) it follows that

$$
a<b<c<A<B<C
$$

and, consequently, the sought product has the form

$$
941 \cdot 852 \cdot 763
$$