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

Question

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 $1/24$

Answer 1

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 $$\small n\equiv 1,11,26,36,41,51,66,76,81,91,106,116,121,131,146,156,161,171,186,196,201,211,226,236\pmod{240}$$

Answer 2

$$\tfrac1{24}n(n+1)(n+2)(n+3)\equiv 1\mod 10\iff n(n+1)(n+2)(n+3)\equiv 24 \mod 240$$ 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: $$\color{red}{n\equiv 16n_1-15n_2\mod 240}.$$

Answer 3

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.