Observe the equation:

$$
\begin{aligned}
(1+2+3+4)^{2} & =(1+2+3+4)(1+2+3+4) \\
& =1 \cdot 1+1 \cdot 2+1 \cdot 3+1 \cdot 4+2 \cdot 1+2 \cdot 2+2 \cdot 3+2 \cdot 4+ \\
& +3 \cdot 1+3 \cdot 2+3 \cdot 3+3 \cdot 4+4 \cdot 1+4 \cdot 2+4 \cdot 3+4 \cdot 4
\end{aligned}
$$

Note that $4 \times 4=16$ products are formed when we calculate $(1+2+3+4)^{2}$ using the distributive property.

a) How many products will be formed when we calculate $(1+2+3+4)^{3}$ also using the distributive property?

b) What is the number of two-digit numbers that use only the digits 1, 2, 3, and 4?
c) What is the sum of the products of the digits of all four-digit numbers formed only by the digits $1, 2, 3,$ and 4?
Solution

a) Each product that appears in the final sum is an expression of the form $x \cdot y \cdot z$ where $x$ is a number from the first parenthesis, $y$ is a number from the second, and $z$ is a number from the third. Since there are 4 possible options for each of these numbers, by the multiplicative principle, we have $4 \cdot 4 \cdot 4=64$ such products.

b) With only 4 options of digits, there are 4 options for the tens digit and four options for the units digit. Thus, we have $4 \cdot 4=16$ two-digit numbers using only the digits 1, 2, 3, and 4.

c) Let's initially calculate the sum of the products of the digits of all two-digit numbers that use only the four given digits. From the previous item, there are 16 such numbers, namely: $11,12,13,14,21,22,23,24,31,32,33,34,41,42,43,44$. The sum of the 16 products of the digits is:

$$
\begin{aligned}
& 1 \cdot 1+1 \cdot 2+1 \cdot 3+1 \cdot 4+2 \cdot 1+2 \cdot 2+2 \cdot 3+2 \cdot 4= \\
& 3 \cdot 1+3 \cdot 2+3 \cdot 3+3 \cdot 4+4 \cdot 1+4 \cdot 2+4 \cdot 3+4 \cdot 4=(1+2+3+4)^{2}
\end{aligned}
$$

This motivates us to analyze the number $(1+2+3+4)^{4}=10000$. Notice that if we apply the distributive property, 256 products will appear, and each of them coincides with the product of the digits of the four-digit numbers formed only by the digits 1, 2, 3, and 4. Thus, the desired value is 10000.