a : 24 = c1 rest 21 ==> a = 24c1 + 21 |+3
a : 15 = c2 rest 12 ==> a = 15c2 + 12 |+3
a : 20 =c3 rest 17 ==> a = 20c3 + 17 |+3
a + 3 = 24(c1+1)
a + 3 = 15(c2+1)
a + 3 = 20(c3+1)
a + 3 = c.m.m.m.c (24; 15; 20)
24 = 2^3 × 3
15 = 3 × 5
20 = 2^2 × 5
--------------------
2^3 × 3 × 5 = 120
a + 3 = 120
a = 120 - 3 = 117 (cel mai mic nr nat care indeplineste conditiile din cerinta)
Verificare:
117 : 24 = 4 rest 21
117 : 15 = 7 rest 12
117 : 20 = 5 rest 17
