<s>[INST] Then sum of digits in $M\times N$ 
If $\bf{M = \underbrace{777777777777777777777}_{99-times}}$ and $\bf{N = \underbrace{999999999999999999999}_{77-times}},$ Then sum of digits in  $M\times N$

$\bf{My\; Try::}$ We can write $$M = 7(1+10+10^2+\cdots \cdots +10^{98}) = \frac{7}{9}(10^{99}-1)$$
and write $$N = 9(1+10+10^2+\cdots \cdots +10^{76}) = \frac{9}{9}(10^{77}-1) = (10^{77}-1)$$
So $$M\times N = \frac{7}{9}(10^{99}-1)\times (10^{77}-1)=\frac{7}{9}\left[10^{196}-10^{77}-10^{99}+1\right]$$
Now how can i solve it after that, Help required, Thanks
 [/INST] Note that $N=10^{77}-1$, hence $MN=$
$\underbrace{7\dots7}_{99}\cdot(10^{77}-1)=$
$\underbrace{7\dots7}_{99}\cdot10^{77}-\underbrace{7\dots7}_{99}=$
$\underbrace{7\dots7}_{99}\underbrace{0\dots0}_{77}-\underbrace{7\dots7}_{99}=$
$\underbrace{7\dots7}_{76}6\underbrace{9\dots9}_{22}\underbrace{2\dots2}_{76}3$
Hence the digit-sum of $MN$ is $7\cdot76+6+9\cdot22+2\cdot76+3=891$.
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