A Huge Number

$\begin{array} {rccc} \underbrace{1}_{\color{#D61F06}1} \times \underbrace{1}_{\color{#D61F06}1} = & \color{#D61F06}1 \\ \underbrace{11}_{\color{#D61F06}2} \times \underbrace{11}_{\color{#D61F06}2} = & 1{\color{#D61F06}2}1 \\ \underbrace{111}_{\color{#D61F06}3} \times \underbrace{111}_{\color{#D61F06}3} = & 12{\color{#D61F06}3}21 \\ \implies \underbrace{111111111}_{\color{#D61F06}9} \times \underbrace{111111111}_{\color{#D61F06}9} = & \boxed{ 12345678{\color{#D61F06}9}87654321} \end{array}$

Proof

Let $n \in \mathbb N$ , $a(n) = \underbrace{111...111}_{\text{\# of 1} = n}$ and $b(n) = \underbrace{123...n...321}_{2n-1 \text{ digits}}$ , where $n$ is the centre digit. And, let us claim that $a^2(n) = b(n)$ for all $n$ and prove it by induction.

• For $n=1$ , $a^2(1) = 1^2 = 1 = b(1)$ , therefore the claim is true for $n=1$ .

• Assuming the claim is true for $n$ , then:

$\begin{aligned} \quad a^2(n+1) & = (10a(n)+1)^2 \\ & = 100a^2(n) + 20a(n) + 1 \\ & = 100b(n) + 20a(n) + 1 \\ & = \underbrace{123...n...32100}_{2n+1 \text{ digits}} + \underbrace{222...2220}_{n+1 \text{ digits}} + 1 \\ & = \underbrace{123...n(n+1)n...321}_{2(n+1)-1 \text{ digits}} \\ & = b(n+1) \end{aligned}$

$\quad$ The claim is also true for $n+1$ , therefore, it is true for all $n$ .

A very simple alternative solution (which is a way, that even a more able 4th grader could follow as we are basically using a version of the column method regarding multiplication; I added the zeroes to the front mainly for the easier tabulation (can be considered as place value, but not necessary):

111,111,111 × 111,111,111= 111,111,111 × 1 + 111,111,111 × 10 + ... + + 111,111,111 × 100,000,000

By using the column method for addition regarding the individual terms, we get:

00,000,000,111,111,111

00,000,001,111,111,110

00,000,011,111,111,100

00,000,111,111,111,000

00,001,111,111,110,000

00,011,111,111,100,000

00,111,111,111,000,000

01,111,111,110,000,000

11,111,111,100,000,000

\text {___________________________}

12,345,678,987,654,321

Hence, our answer should be:

$\boxed {12,345,678,987,654,321}$

Now, we can observe, that we only have one digit "1" in the first and last columns, exactly 2 in the second and second last columns, exactly 3 in the third and third last columns and so on until the first and last nth digits coincide (it is easy to see, that this always happens when squaring an n-digit (all digits are "1") integer).

From this, it is also easy to see, that if the base of the number system is b, and 1≤ n ≤ b -1 (which means, that e.g. n is 9 at most in the decimal, 7 in the octal and 15 in the hexadecimal system), then:

$\overline { \underbrace{11\ldots1}_{n\text{ digits}} } × \overline { \underbrace{11\ldots1}_{n\text{ digits}} } = \overline { 12\ldots(n-1)n(n-1) \ldots 21 }$

There is a theorem which says that , 1 x 1 = 1 , 11 x 11 = 121 , 111 x 111 = 12321 And for 1111 x 1111 = 1234321.

So We see that number of 1s in the number is say n, Then it reaches from 1 to n and again back to 1.

Like this take example of 11111 - 5 ->1's So 1234 $\boxed{5}$ 4321.

Similarly for 111111111 - 9 -> 1's we get,

12345678 $\boxed{9}$ 87654321

Solution: 12345678987654321

Solving the problem completely logically;
$111,111,111 × 111,111,111 = 111,111,111 × (100,000,000 + 10,000,000 + 1,000,000 + \cdots + 10 + 1)$
Multiplying with the first term in bracket we notice it will come 9 $1$ 's followed by 8 $0$ 's.
Next, we observe that as we move on to the second and third and so on towards the last term in paranthesis the number of $1$ 's will remain constant and one of the zeroes will vanish with each step. If we add this new product to our previous total we can observe which of the digits need to be incremented by 1.

We will observe that after each step we leave behind one digit and a new zero is incremented to become 1.
This process is intuitive and hence difficult to explain and seems long but once you understand this you will easily get to know why the beautiful pattern breaks down in case of 10 $1$ 's.
And ofcourse our final answer, it is $\boxed {12345678987654321}$

[ P.S. Typing the answer was like playing a piano :P :) ]

There is a pattern for it. The answer is : 12345678987654321