Solve modular arithmetic equation $\frac{1}{24} \cdot n(n+1)(n+2)(n+3) \equiv 1 \pmod{10}$ From another problem, I have reduced it to: This is the last step in solving:

Solve $\frac{1}{24} \cdot n(n+1)(n+2)(n+3) \equiv 1 \pmod{10}$

How should I begin, a major problem is the $1/24$

You want to find $n$ such that
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{240}\tag1$$
By the way, $$\begin{align}(1)&\Rightarrow n(n+1)(n+2)(n+3)\equiv 4\pmod{10}\\\\&\Rightarrow n\equiv 1,6\pmod{10}\end{align}$$
So, you only need to check if each of the following 48 numbers satisfies $(1)$ :
$$1,11,21,\cdots, 231$$
$$6,16,26,\cdots, 236$$
Hence, the answer is
$$\small n\equiv 1,11,26,36,41,51,66,76,81,91,106,116,121,131,146,156,161,171,186,196,201,211,226,236\pmod{240}$$
