Consideram numerele a,b,c,d,e,f si g
a+b+c+d+e+f+g=119
(a-b)+(a-c)+(a-d)+(a-e)+(a-f)+(a-g)=21
6a-b-c-d-e-f-g=21
Din prima ecuatie:
a=119-b-c-d-e-f-g
Inlocuim in a doua ecuatie :
6 (119-b-c-d-e-f-g)-b-c-d-e-f-g=21
714-7b-7c-7d-7e-7f-7g=21
Impartim ecuatia la 7:
102-b-c-d-e-f-g=3
Luam ecuatia:
6a-b-c-d-e-f-g=21
Putem rescrie prima ecuatie :
120-b-c-d-e-f-g=21
De unde deducem faptul ca
6a=120
a=20
a+b+c+d+e+f+g=119
Rezulta
b+c+d+e+f+g=99
Cele 6 numere nu se pot afla deoarece nu sunt destule date despre ele .
Sper ca te.am ajutat .
