Classic number theory problem

Let the largest number less than 10000 be $x$ and rewriting word problem into system of congruences as $\begin{aligned} & x \equiv 9\mod(21) \\& x\equiv 61\mod(73)\end{aligned}$ Firstly note that moduli are co-prime integers so we can take for some integer $k$ and rewrite the equation as $x = 73k +61\cdots(a)$ Now substituting the value of $x$ to the first congruence we have , $\begin{aligned} 73k +61 \equiv 9\mod(21)\end{aligned}$ Now solving for $k$ we get as $\begin{aligned} k = -1 \mod(21)\end{aligned}$ Writing this congruence for some integer $m$ as below $\begin{aligned} k = 21m -1 \end{aligned}$ Now plugging the value of $k$ in equation (a) $\begin{aligned} x & =73(21m-1)+61 = 1533m - 12 \\& \implies x\equiv -12\mod(1533)\end{aligned}$ Note that $x<10000$ so $\begin{aligned} & 9210 \equiv 0\mod(1533) \\& 9816 \equiv -12\mod(1533)\end{aligned}$ Hence the largest possible of $x<1000$ is $9186$ .

Let the number be $N$ , then:

$\begin{aligned} N & \equiv 61 \text{ (mod 73)} \\ \implies N & \equiv 73n + 61 & \small \color{#3D99F6} \text{where }n \text{ is an integer.} \\ 73n + 61 & \equiv 9 \text{ (mod 21)} \\ 10n - 2 & \equiv 9 \text{ (mod 21)} \end{aligned}$

We note that the smallest positive $n$ that satisfies the congruence is 20 and the larger ones is given by $21m+20$ . Then we have $N = 73(21m+20) + 61 < 10000$ . Implies that the largest $m=5$ and the largest $N = 73(21(5)+20)+61 = \boxed{9186}$ .