Folosim teorema impartirii cu rest.
Notam cu n- numarul cautat si cu c1, c2, c3 - caturile
n=25xc1+24 |+1
n=20xc2+19 |+1
n=12xc3 +11 |+1
Adunam fiecare relatie cu 1:
n+1=25(c1+1)
n+1=20(c2+1)
n+1=12(c3+1)
Atunci n+1=c.m.m.m.c(25, 20, 12)=300 ⇒ n+1 = 300 ⇒ n=299
n : 25 = c₁ rest 24 ⇒ n = 25 × c₁ + 24 l ( + 1 ) ⇒ n + 1 = 25 × ( c₁ + 1 )
n : 20 = c₂ rest 19 ⇒ n = 20 × c₂ + 19 l ( + 1 ) ⇒ n + 1 = 20 × ( c₂ + 1 )
n : 12 = c₃ rest 11 ⇒ n = 12 × c₃ + 11 l ( + 1 ) ⇒ n + 1 = 12 × ( c₃ + 1 )
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⇒ n + 1 = c.m.m.m.c. al numerelor ( 25; 20 si 12 )
25 = 5²
20 = 2² x 5
12 = 2² x 3
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c.m.m.m.c al numerelor ( 25; 20 si 12 ) = 2² x 3 x 5² = 300
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n + 1 = 300 ⇒ n = 300 - 1 ⇒ n = 299 → numarul
Verific:
299 : 25 = 11 rest 24
299 : 20 = 14 rest 19
299 : 12 = 24 rest 11
Raspuns: 299 → numarul
