Question

Determinaţi numerele de forma abcd astfel încât (a − 1) x (b − 2) x (c − 3) x (d + 2019) = 2020.
abcd in baza 10
va rooog ajutați-mă!​

Answer 1

(a-1)(b-2)(c-3)(d+2019)=2020

nu-mi place să explic, ci ca tu să înțelegi logica pe care am folosit-o :)

ptr a forma numărul (abcd) => a,b,c,d∈ℕ*

d=1 , a=2 - constante

|b-2|=1

(b=3)

{b=1}

|c-3|=1

(c=4)

{c=2}

Primul număr

2341=2•10³+3•10²+4•10¹+1•10⁰

Al doilea număr

2121=2•10³+1•10²+2•10¹+1•10⁰

Spor la școală!

Answer 2

$\displaystyle\bf\\\boxed{\bf NU~DESFACEM~PARANTEZELE}~.\\-----------------------\\\\(a-1)(b-2)(c-3)(d+2019)=2020.\\problema~ne~cere~sa~descompunem~numarul~2020~in~produs~de~patru\\factori.\\plecam~de~la~descompunerea~numarului~in~factori~primi.\\dar,~observam~ca~d+2019~este~un~numar~apropiat~de~2020,~deci~evident\\d+2019=2020,~de~unde~\boxed{\bf d=1}~.\\2020(a-1)(b-2)(c-3)=2020 \Leftrightarrow (a-1)(b-2)(c-3)=1.\\1=1\cdot1\cdot1=(-1)(-1)1,~(cu~toate~permutarile).$

$\displaystyle\bf\\\boxed{\bf cazul~1~:~1=1\cdot1\cdot1}~.\\atunci,~a-1=1,~b-2=1~si~c-3=1,~de~unde~a=2,~b=3~si~c=4.\\numarul~este~\boxed{\bf 2341}~.\\in~cazul~2,~a,b,c\in\mathbb{N},~prin~urmare~doar~a-1=1.\\a-1=1,~b-2=-1,~c-3=-1 \implies a=2,~b=1,~c=2.\\numarul~este~\boxed{\bf2121}~.$