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 $1/24$
 A: 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
$$\small n\equiv 1,11,26,36,41,51,66,76,81,91,106,116,121,131,146,156,161,171,186,196,201,211,226,236\pmod{240}$$
A: $$\tfrac1{24}n(n+1)(n+2)(n+3)\equiv 1\mod 10\iff n(n+1)(n+2)(n+3)\equiv 24 \mod 240$$
Using the Chinese remainder theorem, this is equivalent to solving the system of congruences:
$$p(n)=n(n+1)(n+2)(n+3)\equiv\begin{cases}0\mod3 \\
4\mod5\\8\mod 16
\end{cases}$$
The only solution to the second congruence is $\;n\equiv 1\mod 5$, whence the solutions to the first two:
$$\color{red}{n_1\equiv 1,6,7\mod15}. $$
We'll determine the values of the factors and of $p(n)$, according to the value of $n\bmod 16$:
$$\begin{array}{r|cccc||r|cccc}
n\equiv&n+1&n+2&n+3&p(n)&n\equiv&n+1&n+2&n+3&p(n) \\
\hline
 0&1&2&3&0&8&9&10&11&0\\
1&2&3&4&8&9&10&11&12&8\\
2&3&4&5&8&10&11&12&13&8\\
3&4&5&6&8&11&12&13&14&8\\
4&5&6&7&8&12&13&14&15&8\\
5&6&7&8&0&13&14&15&0&0\\
6&7&8&9&0&14&15&0&1&0\\
7&8&9&10&0&15&0&1&2&0
\end{array}$$
Thus the solutions of the second congruence are $\;\color{red}{n_2\equiv 1,2,3,4,9,10,11,12\mod16}$.
Byy the inverse isomorphism of the Chinese remainder theorem, from the Bézout's relation: $\;16-15=1$, we obtain the $24$ solutions:
$$\color{red}{n\equiv 16n_1-15n_2\mod 240}.$$
A: Okay enough of this mod $240$ nonsense, the prime $3$ is irrelevant and you have an extra power of $2$ for some reason. As someone in the comments said, this equation is just ${n+3 \choose 4} \equiv 1 \mod 10$.  
Now a nice fact to know is that as a function of $n$ the binomial coefficients $n \choose k$ are periodic modulo any prime $p$ with period $p^ {\lceil log_p(k+1)\rceil}$. If you want to know why this is true, look up Lucas' theorem on binomial coeffients.
So for $k=4$ this means the function ${n+3 \choose 4}$ is periodic modulo $2$ with period $8$, and periodic mod $5$ with period $5$. Solving these separately we see $n \equiv 1,2,3,4 \mod 8$ and $n \equiv 1 \mod 5$ giving solutions $n \equiv 1,11,26,36 \mod 40$. This gives the same answers as mathlove just less redundant.
