Factorials/102

Since $102 = 2 \times 3 \times 17$ , $n! \bmod 102 = 0$ for all $n \ge 17$ . Therefore we have:

$\begin{aligned} \sum_{n=1}^{100} n! & \equiv 1! + 2! + 3! + 4! + 5! + \cdots + 16! \text{ (mod 102)} \\ & = 1 + 2 + 6 + 24 + (4!)5 + \cdots + 16! \text{ (mod 102)} \\ & = 33+ (24)5 + (5!)6 + \cdots + 16! \text{ (mod 102)} \\ & = 33+ 18 + (18)6 + \cdots + 16! \text{ (mod 102)} \\ & = 51 + 6 + (6)7 + \cdots + 16! \text{ (mod 102)} \\ & = 57 + 42 + (42)8 + \cdots + 16! \text{ (mod 102)} \\ & = 99 + 30 + (30)9 + \cdots + 16! \text{ (mod 102)} \\ & = 27 + 66 + (66)10 + \cdots + 16! \text{ (mod 102)} \\ & = 93 + 48 + (48)11 + \cdots + 16! \text{ (mod 102)} \\ & = 39 + 18 + (18)12 + \cdots + 16! \text{ (mod 102)} \\ & = 57 + 12 + (12)13 + 14! + 15! + 16! \text{ (mod 102)} \\ & = 69 + 54 + (54)14 + 15! + 16! \text{ (mod 102)} \\ & = 21 + 42 + (42)15 + 16! \text{ (mod 102)} \\ & = 81 + 84 \text{ (mod 102)} \\ & = \boxed{63} \text{ (mod 102)} \end{aligned}$

$\begin{array}{r|l} 102&2\\ 51&3\\ 17&17\\ 1 \end{array}$

Therefore after 16 each factorials will be divisible by 102.

Let's start to calculate the remainder for each factorials: $\begin{array}{c|c|c} \text{number}&\text{calculation}&\text{remainder}\\ \hline 1&1\\ 2&1\times2&2\\ 3&2\times3&6\\ 4&6\times4&24\\ 5&24\times5\pmod{102}&18\\ 6&18\times6&108\\ \vdots \end{array}$ If you continue that: $\displaystyle \sum_{n=1}^{16}n!\pmod{102}=63$

anyone please write the solution