Our Year In A Big Number.

Each block contains the sum of all the numbers from 1 to 9: 45.

Then, 2018 = 45 × 44 + 38.

So now, we just have to find the 38th number of a block, which is 9.

• In a single block ( ${\color{#D61F06}{122333\ldots999999999}}$ ), there are $1 + 2 + \ldots 9 \implies \frac{9\cdot10}{2} \implies \color{#20A900}{45}$ digits.
• $2018 = {\color{#D61F06}{45}}\cdot 44 + \color{#20A900}{38}$ means essentially counting the 38th number which is $\boxed{9}$

$\underbrace 1_{T_1}2 \underbrace 2_{T_2}3\ \ 3\underbrace 3_{T_3}4\ \ 4\ \ 4\underbrace 4_{T_4}\cdots\ \ 9\ \ 9\ \ 9\ \ 9\ \ 9\ \ 9\ \ 9\ \ 9\underbrace 9_{T_9}$

Let the digit 1 to 9 be $n$ . We note that the $k$ th digit "position" of the last digit $n$ is given by triangle number $T_n = \frac {n(n+1)}2$ . For example, the last digit 1 is at $T_1 = \frac {1(1+1)}2 = 1$ ; the last digit 2 is at $T_2 = \frac {2(2+1)}2 = 3$ ; the last digit 4 is at $T_4 = \frac {4(4+1)}2 = 10$ ; ... the last digit 9 is at $T_9 = \frac {9(8+1)}2 = 45$ . The length of one block is therefore $T_9$ or 45. Therefore the positive of the 2018th digit in a block is given by $2018 \bmod 45 = 38$ . Since $38 > T_{\color{#D61F06}8} = \frac {8(8+1)}2 = 36$ , the 2018th digit must be after $\color{#D61F06}8$ which is $\boxed{9}$ .