Question

Doar pentru pasionati .

Answer 1

$\displaystyle\\(a,b,c,\underline{d})\in\mathbb{N}^4~astfel~incat~2\cdot3^a+8\cdot3^b+5\cdot3^c+3^d=2022.~~(1)\\-----------------------------------\\O~prima~observatie~care~"sare-n~ochi"~este~ca~avem~multe~puteri~\\\\ale~lui~3~in~membrul~stang~ceea~ce~sugereaza~sa~folosim~baza~de~\\\\numeratie~3.~Astfel,~2022=2202220_{(3)}=2\cdot3+2\cdot3^2+2\cdot3^3+2\cdot3^5+2\cdot3^6.\\\\Acum,~rescriem~membrul~stang~al~relatiei~(1)~convenabil,~astfel,~obtinem$

$\displaystyle\\2\cdot3^a+8\cdot3^b+5\cdot3^c+3^d=2\cdot3^a+2\cdot3^b+2\cdot3^{b+1}+2\cdot3^c+3^{c+1}+3^d.\\\\Prin~urmare,~2\cdot3^a+2\cdot3^b+2\cdot3^{b+1}+2\cdot3^c+3^{c+1}+3^d=2\cdot3+2\cdot3^2+2\cdot3^3+2\cdot3^5+2\cdot3^6,\\\\si~observatia~"cruciala"~in~acest~caz~este~ca~in~ambii~membri~avem~termeni\\\\puteri~ale~lui~3,~insa,~in~membrul~stang~si~drept~avem~5~respectiv~6~astfel\\\\de~termeni,~si~mai~mult~in~membrul~drept~toti~acesti~astfel~de~termeni~sunt$

$\displaystyle\\inmultiti~cu~2~pe~cand~in~membrul~stang~doar~4~sunt~inmultiti,~prin~urmare\\\\trebuie~sa~avem~\boxed{c+1=d}.~Astfel,~relatia~(1)~in~"forma~finala"~se~scrie~\\\\2\cdot3^a+2\cdot3^b+2\cdot3^{b+1}+2\cdot3^c+2\cdot3^{c+1}=2\cdot3+2\cdot3^2+2\cdot3^3+2\cdot3^5+2\cdot3^6,\\\\si~se~observa~ca~\boxed{(a,b,c,d)\in\left\{(1,2,5,6),(1,5,2,3),(3,5,1,2),(3,1,5,6)\right\}}.\\-----------------------------------$

$\displaystyle\\(*)~Observatie:~In~enunt~este~o~eroare,~se~specifica~doar~a,b,c\in\mathbb{N},~dar\\si~d\in\mathbb{N},~acest~lucru~demonstrandu-se~usor~deoarece~din~relatia~(1)~se\\obtine~3^d=2022-(2\cdot3^a+8\cdot3^b+5\cdot3^c)\in\mathbb{N}~iar~acest~lucru~impune~evident\\si~d\in\mathbb{N}.$