Question

vious Years NTSE Questions
If m = 7777....7777 is a 99 digit number and
n = 999....999 is 77 digit number, then the sum of
the digits in the product mx n is
(1) 890
(2) 891
(3) 892
(4) 893
(Odisha, Stage-1, 2012-13)​

Answer 1

It's just simple!

Here 'm' is given as 777......777 having 99 digits.

$m=\underbrace{777\dots\dots 777}_{99}$

And 'n' is given as 999......999 having 77 digits.

$n=\underbrace{999\dots\dots999}_{77}$

All we have to do is just to write 'n' in connection with power of 10 !

$n=\underbrace{999\dots\dots999}_{77}=1\underbrace{000\dots\dots000}_{77}-1=10^{77}-1$

Let's begin!

$\begin{aligned}&mn\\ \\ \Longrightarrow\ \ &\underbrace{777\dots\dots777}_{99}\times \underbrace{999\dots\dots999}_{77}\\ \\ \Longrightarrow\ \ &\underbrace{777\dots\dots777}_{99}\ (10^{77}-1)\\ \\ \Longrightarrow\ \ &\underbrace{777\dots\dots777}_{99}\underbrace{000\dots\dots000}_{77}\ -\ \underbrace{777\dots\dots777}_{99}\end{aligned}$

Here we might have a thought.

Consider the number  $\underbrace{777\dots\dots777}_{99}\underbrace{000\dots\dots000}_{77}.$  The other number is subtracted from this number. We call this number as the minuend.

After the subtraction, in the minuend,

→  The last digit 0 borrows from its left to be 10, thus it becomes 10 - 7 = 3.

→  The other 76 zeroes each borrow from their lefts to be 10, then each are borrowed towards their rights being 9. Thus these digits become 9 - 7 = 2.

→  Since 'm' has 22 digits more than 'n', the 22 sevens each at the right of those 7s from which subtraction occurs, are borrowed towards their rights to be 6, then each borrows from their lefts becoming 16. Thus they each become 16 - 7 = 9.

→  As a result of this process, the 77th seven becomes 6, because this guy is borrowed towards its right.

→  And the other 76 sevens remain unchanged.

If feel trouble understanding the written, have a look at the basic subtraction method given below.

$\begin{aligned}&\overbrace{777\dots\dots777}^{77}\!\!\!\!\!&\overbrace{777\dots\dots777}^{22}\overbrace{000\dots\dots000}^{77}\\ &-&777\dots\dots777777\dots\dots777\\ \cline{1-3}&\underbrace{777\dots\dots77}_{76}6\!\!\!\!\!&\underbrace{999\dots\dots999}_{22}\underbrace{222\dots\dots22}_{76}3\end{aligned}$

The resultant is the value of 'mn' !

So, the sum of digits in 'mn' is,

$7\cdot 76+6+9\cdot 22+2\cdot 76+3=\bold{891}$

Hence option (2) is the answer.

Answer 2

Hey there

refer to attachment