Problem - Sum of digits

Question

For every numbers with $3$ digits you calculate the product of the numbers. After that you take the sum of the products, what number do you get?

I didn't know how to do this exactly. What would be the easiest way?

Answer 1

Suppose you do this for 1-digit numbers, then the answer is $0+1+2+...+9=45$.

Suppose you do this for 2-digit numbers, then the answer is \begin{align} & 0\cdot 0 + 0\cdot1+\cdots+0\cdot9 + \\ & 1\cdot 0 + 1\cdot1+\cdots+1\cdot9 + \\ & \quad\cdots\quad\cdots\quad\cdots\quad\cdots \\ & 9\cdot0 + 9\cdot1+\cdots+9\cdot9 \end{align} which factors as $(0+1+\cdots+9)^2$.

Suppose you do this for 3-digit numbers, then the answer will analogously factor as $(0+1+\cdots+9)^3$.

In general, for $k$-digit numbers, the answer is $45^k$.

Answer 2

Sum of n1*n2*n3 with 0 <= n1,n2,n3 <= 9 = (Sum of n1 with 0 <= n1 <= 9) * (Sum of n2 with 0 <= n2 <= 9) * (Sum of n3 with 0 <= n3 <= 9) = (Sum of n with 0 <= n <= 9)^3 = (9*10/2)^3 = 45^3 = 91125

Answer 3

Your sum factorises to $(1+2+\ldots+9)(1+2+\ldots+9)(1+2+\ldots+9)=45^{3}$.
To see this, expand term by term.