Galperin G.A.

a) For each three-digit number, we take the product of its digits, and then we add up these products, calculated for all three-digit numbers. What will be the result?

b) The same question for four-digit numbers.
a) $(1+2+\ldots+9)(0+1+\ldots+9)(0+1+\ldots+9)=45^{3}$, since the sum in each parenthesis is 45. On the other hand, if we expand the parentheses on the left side, in the resulting sum, there will be products corresponding to all three-digit numbers (for example, the product $2 \cdot 5 \cdot 7$, corresponding to the number 257, will appear when we take the two from the first parenthesis, the five from the second, and the seven from the third).

b) It is solved similarly.

## Answer

a) $45^{3} ;$ b) $45^{4}$.