[tex]\displaystyle\bf\\\boxed{\bf NU~DESFACEM~PARANTEZELE}~.\\-----------------------\\\\(a-1)(b-2)(c-3)(d+2019)=2020.\\problema~ne~cere~sa~descompunem~numarul~2020~in~produs~de~patru\\factori.\\plecam~de~la~descompunerea~numarului~in~factori~primi.\\dar,~observam~ca~d+2019~este~un~numar~apropiat~de~2020,~deci~evident\\d+2019=2020,~de~unde~\boxed{\bf d=1}~.\\2020(a-1)(b-2)(c-3)=2020 \Leftrightarrow (a-1)(b-2)(c-3)=1.\\1=1\cdot1\cdot1=(-1)(-1)1,~(cu~toate~permutarile).\\[/tex]
[tex]\displaystyle\bf\\\boxed{\bf cazul~1~:~1=1\cdot1\cdot1}~.\\atunci,~a-1=1,~b-2=1~si~c-3=1,~de~unde~a=2,~b=3~si~c=4.\\numarul~este~\boxed{\bf 2341}~.\\in~cazul~2,~a,b,c\in\mathbb{N},~prin~urmare~doar~a-1=1.\\a-1=1,~b-2=-1,~c-3=-1 \implies a=2,~b=1,~c=2.\\numarul~este~\boxed{\bf2121}~.[/tex]
