Divisor of factorial number

The first step is to determine the number of divisors of $10!$ . Then, determine the number of divisors of the original number $\bmod\, 210$ .

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Recall that De Polignac's formula is

$s_p(n) = \sum\limits_{j=1}^{\lfloor \log_p(n) \rfloor} \left\lfloor \dfrac{n}{p^j}\right\rfloor$ where $p$ is a positive prime integer.

Let $n = 10$ . Since there are $4$ positive prime integers between $2$ and $10$ , namely $2,3,5$ and $7$ ,

• If $p_1 = 2$ , then $s_{p_1}(n) = 8$ ;
• If $p_2 = 3$ , then $s_{p_2}(n) = 4$ ;
• If $p_3 = 5$ , then $s_{p_3}(n) = 2$ ;
• If $p_4 = 7$ , then $s_{p_4}(n) = 1$ ;

which illustrates the sequence of decreasing number of divisors for increasing prime numbers. We can then express $(10!)^{10!}$ as $\left(10!\right)^{10!} = \left(2^8 \cdot 3^4 \cdot 5^2 \cdot 7\right)^{10!} = 2^{8 \cdot 10!} \cdot 3^{4 \cdot 10!} \cdot 5^{2 \cdot 10!} \cdot 7^{10!}$

Looking at the exponents of the divisors, we see that the total number of divisors is $\left(8 \cdot 10! + 1\right)\left(4 \cdot 10! + 1\right)\left(2 \cdot 10! + 1\right)\left(\cdot 10! + 1\right)$ Notice that since $210 = 2 \cdot 3 \cdot 5 \cdot 7$ and $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$ , this shows that $210 | 10!$ . In that case, $10! \equiv 0 \bmod 210$ . Thus, $\left(8 \cdot 10! + 1\right)\left(4 \cdot 10! + 1\right)\left(2 \cdot 10! + 1\right)\left(\cdot 10! + 1\right) \equiv \boxed{1 \bmod 210}$

Notice, $(10!)^{10!}$ = $(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9\cdot 10)^{10!}$ =( $2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7)^{10!}$ = $2^{8\cdot 10!}\cdot 3^{4\cdot 10!}\cdot 5^{2\cdot 10!}\cdot 7^{10!}$ .So,number of divisor of $(10!)^{10!}$ is = $(8 \cdot 10!+1)$ $\cdot$ $(4 \cdot 10!+1)$ $\cdot$ $(2 \cdot 10!+1)$ $\cdot$ $(10!+1)$ . Now look, $210 | 8\cdot 10!$ ;so, $(8\cdot 10!+1)$ $\equiv$ 1 $\pmod{210}$ ------------------- $(1)$ ; $210 | 4\cdot 10!$ ;so, $(4\cdot 10!+1)$ $\equiv$ 1 $\pmod{210}$ ------------------- $(2)$ ; $210 | 2\cdot 10!$ ;so, $(2\cdot 10!+1)$ $\equiv$ 1 $\pmod{210}$ ------------------- $(3)$ ; $210 | 10!$ ;so, $(10!+1)$ $\equiv$ 1 $\pmod{210}$ ------------------- $(4)$ ; $(1)$ , $(2)$ , $(3)$ & $(4)$ implies, $(8\cdot 10!+1)\cdot (4\cdot 10!+1)\cdot (2\cdot 10!+1)\cdot (10!+1)$ $\equiv$ 1 $\pmod{210}$ . So,the desired ans is $1$ .