a + b + c + d = 120 → suma celor patru numere
a - 5 = b : 2 = 3 × c = 2 × d = k
a - 5 = k ⇒ a = k + 5
b : 2 = k ⇒ b = 2 k
3 × c = k ⇒ c = k/3
2 × d = k ⇒ d = k/2
→ inlocuim in suma
( k + 5) + ( 2 k ) + ( k/3 ) + ( k/2) = 120
↓ ↓ ↓ ↓
a + b + c + d = 120
3 k + k/3 + k/2 = 120 - 5 = 115 → le aduc la acelasi numitor comun 6
6× 3 k + 2 k + 3 k = 6 × 115
23 × k = 690
k = 690 : 23
k = 30
a - 5 = 30 ⇒ a = 30 + 5 = 35 → primul numar
b = 2 × 30 = 60 → al doilea numar
c = 30 : 3 = 10 → al treilea numar
d = 30 : 2 = 15 → al patrulea numar
Verific: 35 + 60 + 10 + 15 = 120 → suma celor patru numere
