Question Number 188359 by 073 last updated on 28/Feb/23
Answered by Rasheed.Sindhi last updated on 28/Feb/23
$$\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+...+\mathrm{48}!\equiv{x}\left({mod}\:\mathrm{6}\right) \\ $$$${For}\:{k}\geqslant\mathrm{3},\:\mathrm{6}\mid{k}!\: \\ $$$$\Rightarrow\mathrm{1}!+\mathrm{2}!\equiv{x}\left({mod}\:\mathrm{6}\right) \\ $$$$\:\:\:\:\:\mathrm{1}+\mathrm{2}\equiv{x}\left({mod}\:\mathrm{6}\right. \\ $$$${x}=\mathrm{3} \\ $$
Commented by 073 last updated on 01/Mar/23
$$\mathrm{thanks}\:\mathrm{alot} \\ $$