Find The Pattern - 3

We note that the pattern is $\underbrace{123 \cdots}_{n-1 \text{ digits}} \times 9 + n = \underbrace{111 \cdots}_{n \text{ digits}}$ for $n \ge 2$ . Therefore, for $n = 10$ , we have $\boxed{\underbrace{1111111111}_{10 \ \ 1 \text{'s}}}$ .

For what is happening:

Let the general equation be as follows, where positive integer $n<10$ .

$\begin{aligned} \underbrace{\overline{123...n}}_{n \text{ digits}} \times 9 +n+1 & = \overline{123...n} \times (10-1) + n + 1 \\ & = \underbrace{\overline{123...n0}}_{n+1 \text{ digits}} - \underbrace{\overline{123...n}}_{n \text{ digits}} + \color{#3D99F6}{n + 1} \quad \quad \small \color{#3D99F6}{\text{Let }p=n+1} \\ & = \overline{123...n\color{#3D99F6}{p}} - \overline{123...n} \\ & = \underbrace{\overline{111...1}}_{n+1 \text{ digits}} \end{aligned}$

Let $n$ be a single digit positive integer . And denote $S_n$ as the $n$ -digit integer, $\overline{123\ldots n}$ , that is $S_n$ is concatenation of the first $n$ positive integers in ascending order.

The given list of equations shows that $9S_n + (n+1) = \underbrace{1111111\ldots1}_{(n+1) \text{ number of 1's}}$ for $n=1,2,\ldots, 8$ .

And we want to determine the value of $9S_9 + 9 + (9 + 1)$ , so a natural question to ask is: "Does the pattern above also hold true for $n= 9$ ?" Let's find out.

Claim : The following equation holds true for $n=1,2,\ldots, 9$ .

$9S_n + (n+1) = \underbrace{1111111\ldots1}_{(n+1) \text{ number of 1's}} \qquad (\bigstar)$

Proof : If we add $S_n$ to both sides of equation $(\bigstar)$ and rearrange the terms, we get

$10S_n = \left( \underbrace{1111111\ldots1}_{(n+1) \text{ number of 1's}} + S_n \right) - (n+1) \qquad (\bigstar\bigstar)$

In RHS, let us try to simplify the expression inside the bracket, note that $S_n$ is an $n$ -digit number and $\underbrace{1111111\ldots1}_{(n+1) \text{ number of 1's}}$ is an $(n+1)$ -digit number. We have

$\large{\begin{array}{cccccccc} && & 1 & 1& 1 & \ldots &1&1\\ +&& & & 1& 2 & \ldots& (n-1)& n \\ \hline & & & 1 & 2& 3 &\ldots & n&(n+1)\\ \hline \end{array}}$

Looking at the long addition above shows that $\underbrace{1111111\ldots1}_{(n+1) \text{ number of 1's}} + S_n = \overline{123\ldots (n)(n+1)}$ . If we subtract it by $(n+1)$ , we get

$\begin{aligned} RHS &=& \overline{123\ldots n0} = 10S_n = LHS. \end{aligned}$

This shows that our claim is true for all single digit positive integer $n$ .

Hence, $9S_9 + 10 = 123456789 \times 9 + 10$ is indeed equal to $\underbrace{1111111\ldots1}_{10 \text{ number of 1's}}$ .

When+9then ans111111111 So+10then ans 1111111111

The value we have to find is the pattern which continues the tower shown in the question. We observe that a $\color{#3D99F6}{1}$ is consecutively added to every answer. The last answer in the tower have nine $\color{#3D99F6}{1}$ 's. So, the next one should have one more $\color{#3D99F6}{1}$ that is ten $\color{#3D99F6}{1}$ 's. So, our answer is $\color{#D61F06}{\boxed{1111111111}}$ .