B3. 18126 A good strategy is to first look at small examples. This way, we find successively

$$
\begin{array}{rccc}
9 \times 4 & = & 40-4 & =36, \\
99 \times 44 & = & 4400-44 & =4356, \\
999 \times 444 & = & 444000-444 & =443556, \\
9999 \times 4444 & = & 44440000-4444 & =44435556 .
\end{array}
$$

The pattern will be clear. To answer the question, we note that $999 \ldots 99 = 1000 \ldots 00 - 1$, where the first number has 2014 zeros. The product is therefore equal to

$$
\underbrace{444 \ldots 44}_{2014 \text { fours }} \underbrace{000 \ldots 00}_{2014 \text { zeros }} - \underbrace{444 \ldots 44}_{2014 \text { fours }} = \underbrace{444 \ldots 44}_{2013 \text { fours }} 3 \underbrace{555 \ldots 55}_{2013 \text { fives }} 6 .
$$

If we add the digits of this number, we get $2013 \cdot 4 + 3 + 2013 \cdot 5 + 6 = 2013 \cdot 9 + 9 = 18126$.
B3. If you were to calculate the result of

$$
\underbrace{999 \ldots 99}_{2014 \text { nines }} \times \underbrace{444 \ldots 44}_{2014 \text { fours }}
$$

and then add up all the digits of the result, what would the outcome be?