[tex]\displaystyle Fie~a,b,c,d,e~numerele. \\ \\ Avem:~ \frac{a}{4}= \frac{b}{5}= \frac{c}{6} ~si~ 4c=5d=6e. \\ \\ Impartind ~a~doua~relatie~prin~24,~obtinem: \\ \\ \frac{c}{6}= \frac{5d}{24} = \frac{e}{4},~si~tinand~cont~de~prima~relatie,~avem: \\ \\ \frac{a}{4}= \frac{b}{5}= \frac{c}{6}= \frac{5d}{24}= \frac{e}{4}. ~Inmultind~aceasta~relatie~cu~120, \\ \\ obtinem: ~ \boxed{30a=24b=20c=25d=30e=k}. [/tex]
[tex]\displaystyle k ~reprezinta~rezultatul~comun~al~acelor~numere,~iar \\ \\ pentru~ca~acesta~sa~fie~minim~(deoarece~cifrele~sunt~nenule), \\ \\ el ~trebuie~sa~fie~egal~cu~c.m.m.m.c~al~numerelor~30,~24,~20, \\ \\ 25,~30. \\ \\ ~ [30,24,20,25,30]=600. \\ \\ k=600 \Rightarrow a=20,~b=25,~c=30,~d=24,~e=20. \\ \\ Acestea~sunt~cele~mai~mici~numere~cu~proprietatea~din~enunt. \\ \\ Evident,~solutia~generala~este~de~forma~ \\ \\ (a,b,c,d,e)=(20t,~25t,~30t,~24t,~20t). [/tex]
[tex]\displaystyle t \in N^*. \\ \\ Pentru~punctul~b)~avem: \\ \\ 20t+25t+30t+24t+20t=357 \Leftrightarrow 119t=357 \Rightarrow t=3. \\ \\ Solutia~pentru~punctul~b)~este: \\ \\ (a,b,c,d,e)=(60,75,90,72,90).[/tex]
