Multiplication And Addition Are Related!

Let's first evaluate $11×11 = 121$
Then $111×111= 12321$
Then $1111×1111=1234321$

Let $n$ be the number of digits of the question. So, in each case we observe that the digits of the answer is of the form:-
$1,2,3, \cdots, n, (n-1), (n-2), \cdots , 3,2,1$

In this case $n = 9$
$\therefore$ The answer will be $12345678987654321$ . $_\square$

Every $111\dots111$ with " $n$ " 1's $=10^{n-1}+10^{n-2} + \dots + 100+10+1$ . Using the formula for GP sum , we have the above expression as $\dfrac{1(10^{n}-1)}{10-1}=\dfrac{10^n-1}{9}$ .

Thus,

$111\dots111 \times 111\dots111 = \left(\dfrac{10^n-1}{9}\right)^2 = \dfrac{10^{2n}-2\times10^n+1}{81} = \dfrac{10^n(10^n-2)+1}{81}$

Thus when (in this problem) $n=9$ , we have:

$\dfrac{10^n(10^n-2)+1}{81} = \dfrac{10^9(999999998)+1}{81} = \dfrac{999999998000000001}{9\times 9}$

$=12345678987654321$

Note: We have reduced the tedious multiplication to simple division.

When we try out 11x11 and 111x111 and 1111x1111 we mostly understand the pattern of this question .