The integers from 1 to $k$ are concatenated to form the integer $N=123456789101112 \ldots$ Determine the smallest integer value of $k>2019$ such that $N$ is divisible by 9 .
We can write the integer $N$ as

$$
N=1 \cdot 10^{a_{1}}+2 \cdot 10^{a_{2}}+3 \cdot 10^{a_{3}}+\cdots+(k-1) \cdot 10^{a_{k-1}}+k \cdot 10^{0}
$$

for some positive integers $a_{1}, a_{2}, a_{3}, \ldots a_{k-1}$.

Suppose that

$$
M=1 \cdot\left(10^{a_{1}}-1\right)+2 \cdot\left(10^{a_{2}}-1\right)+3 \cdot\left(10^{a_{3}}-1\right)+\cdots+(k-1) \cdot\left(10^{a_{k-1}}-1\right)
$$

Since the digits of $10^{a_{j}}-1$ are all $9 \mathrm{~s}$ for each $j$ with $1 \leq j \leq k-1$, then each $10^{a_{j}}-1$ is divisible by 9 .

This means that $M$ is divisible by 9 .

Since $M$ is divisible by 9 , then $N$ is divisible by 9 exactly when $N-M$ is divisible by 9 .

But

$N-M=1 \cdot\left(10^{a_{1}}-\left(10^{a_{1}}-1\right)\right)+2 \cdot\left(10^{a_{2}}-\left(10^{a_{2}}-1\right)\right)+\cdots+(k-1) \cdot\left(10^{a_{k-1}}-\left(10^{a_{k-1}}-1\right)\right)+k \cdot 10^{0}$

and so

$$
N-M=1+2+\cdots+(k-1)+k=\frac{1}{2} k(k+1)
$$

Therfore, $N$ is divsible by 9 exactly when $\frac{1}{2} k(k+1)$ is divisible by 9 .

Since $k>2019$, we start checking values of $k$ at $k=2020$.

When $k=2020,2021,2022,2023$, the value of of $\frac{1}{2} k(k+1)$ is not divisible by 9 .

When $k=2024$, we see that $\frac{1}{2} k(k+1)=\frac{1}{2}(2024)(2025)=1012 \cdot 9 \cdot 225$.

Therefore, $k=2024$ is the smallest $k>2019$ for which $N$ is divisible by 9 .

ANSWER: 2024