n × (n + 1) ∈ D₃₆₀
=> n ∈ D₃₆₀ si (n + 1) ∈ D₃₆₀
unde n si n+1 sunt numere consecutive.
Descompunem numarul 360
360 = 6 × 6 × 10 = 2³ × 3² × 5
Scriem multimea divizorilor lui 360, in care cautam numere consecutive:
D₃₆₀ = {1; 2; 3; 4; 5; 6; 8; 9; 10; 12; 15; 18; 20; 24; 30; 36; 40; 45; 60; 72; 90; 120; 180; 360}
Avem mai multe solutii:
S1: 1 si 2; 1 × 2 = 2; 2 ∈ D₃₆₀; => n = 1
S2: 2 si 3; 2 × 3 = 6; 6 ∈ D₃₆₀; => n = 2
S3: 3 si 4; 3 × 4 = 12; 12 ∈ D₃₆₀; => n = 3
S4: 4 si 5; 4 × 5 = 20; 20 ∈ D₃₆₀; => n = 4
S5: 5 si 6; 5 × 6 = 30; 30 ∈ D₃₆₀; => n = 5
S6: 8 si 9; 8 × 9 = 72; 72 ∈ D₃₆₀; => n = 8
S7: 9 si 10; 9 × 10 = 90; 90 ∈ D₃₆₀; => n = 9
In total, problema are 7 solutii.
