Advanced Modular Arithmetic #1

Note that $26!, 27!, 28!, \cdots$ can be divided by $13$ twice or more.

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Simplify the given equation and we get

$a=n!\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{22}+\frac{1}{23}\right)$

Because $a$ must be an integer, $n$ must be larger than or equal to $23$ .

Since $\frac{n!}{k}$ can be divided by $13$ given that $k\neq13$ ,

$a \pmod{13} \equiv \frac{n!}{13} \pmod{13}$ .

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If $n\geq26$ , as said before, $\frac{n!}{13^2}$ is a natural number, which makes $\frac{n!}{13} \pmod{13}\equiv0 \pmod{13}$ .

Then $a \pmod{13}\equiv0\pmod{13}$ , but this is exactly the opposite of what the question wants.

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Therefore the maximum of $n$ is $25$ . $\Rightarrow \boxed{M=25}$ .

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According to the question, we're trying to get the value of $R$ when $n=M-1=24$ .

$\begin{aligned} a \pmod{13} & \equiv \frac{24!}{13} \pmod{13} \\ & \equiv 1\cdot2\cdot3\cdot4\cdots11\cdot12\cdot(13+1)\cdot(13+2)\cdot(13+3)\cdot(13+4)\cdots(13+10)\cdot(13+11) \pmod{13} \\ & \equiv (1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11)^2\cdot12 \pmod{13} \\ & \equiv \{(2\cdot7)\cdot(3\cdot9)\cdot(4\cdot10)\cdot(5\cdot8)\cdot(6\cdot11)\}^2\cdot12 \pmod{13} \\ & \equiv \{(13+1)\cdot(2\cdot13+1)\cdot(3\cdot13+1)\cdot(3\cdot13+1)\cdot(5\cdot13+1)\}^2\cdot12 \pmod{13} \\ & \equiv 12 \pmod{13} \end{aligned}$

$\therefore \boxed{R=12}$

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$\therefore \boxed{M\cdot R=300}$