Find the sum of digits in the product of $A$ and $B$ My question is
If $A$= $9999...........$($77$ times or $77$ digits) and 
$B$= $7777............$($99$ times or $99$ digits)
Then find the sum of digits in the product of $A$ and $B$.
I have a query that if we have same number of digits then i can easily figure out the sum but in the question it is different. Any help will be appreciated.
Note: $$9.7 = 63$$
      $$99.77=7623$$
      $$999.777=776223$$ and so on...
 A: $A=999...(77$ times)= $10^{78}-1$
$B=777...(99$ times)= $10^{98}.7+10^{97}.7+.....+10^{77}+10^{76}.7+10^{75}.7+....+7$
so.
$AB=10^{78}.B-B$
$= 77..(78$times)$6$$22...(97$times)3
$\implies$ sum of the digits= $78 \times 7 + 6 + 97 \times 2 +3  $ 
there can be some mistake in the calculation. but i think we can do it this way
A: we have $A=999...9$ which has $77$ digits and  $B=777..7$ which has $99$ digits.
Now Let $X=777..7$ which has $77$ digits and $Y=777..7000...0$ which has $22$ sevens and $77$ Zeros. So
$$B=777..7+777..7000...0=X+Y$$ Also $Y=Z*10^{77}$ where $Z=777..77$ with $22$ Sevens. 
$$AB=A(X+Y)$$
According to your Note $$AX=777..76222...23$$ which has $76$ Sevens and $76$ Twos.So Total digits of number $AX$ is $154$.
Similarly $$AY=0000..0777..7000..0$$ which has $55$ Zeros in front, $22$ Sevens and $77$ Zeros at last with total digits as again $154$. Now simply add $AX$ and $AY$ by aligning $154$ digits of $AX$ with corresponding $154$ digits of $AY$ you will get the answer.
A: In light  of@user300 answer
The resultant number will be 
$ \quad \underbrace {77....777}_{76digits}\;6\quad \underbrace {99....999}_{22{\kern 1pt} digits}\;\underbrace {22....222}_{76{\kern 1pt} digits}3$
So the sum =$76 \times 7 +22 \times 9+76 \times 2 +6+3=891$
