Answer

The answer above is correct, because $\alpha = 10^{2010} - 1$ and so $\gamma = \overline{4\ldots 435\ldots 56}$, so the sum of its digits is $2010\cdot 4 - 1 + 2010\cdot 5 + 1 = 2010\cdot 9 = 18090$.

MathSolver94 wrote: Let $\gamma=\alpha \times \beta$ where \[\alpha=999 \cdots 9\] (2010 '9') and \[\beta=444 \cdots 4\] (2010 '4') Find the sum of digits of $\gamma$. I can imagine someone guessing the answer on the test like this: $9\times 4 = 36$ $99\times 44 = 4356$ $999\times 444=443556$ ...

$99999....9*A=10^kA-A=(A-1)*10^k+1+(9....9-A)$. Therefore $s(9...9*A)=9*k$ for any $A<10^k$.

$\gamma = 9(111...11) \times 4(111...11)$ $\gamma = 36(111...11)$ $\gamma = 3999...996$ with 2009 $9$'s so the sum of the digits is $3 + 9 \times 2009 + 6 = 18090$ even though the gamma is false but the sum of digits is still somehow correct. can someone tell me why?