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aflati cel mai mic numar natural care impartit la 21,35si42,da resturile18,32si respectiv39

Răspuns :

D=Î×C+R
D=21×C1+18|+3=>D+3€M21 》
D=35×C2+32|+3=>D+3€ M35 》=>
D=42×C3+39|+3=>D+3 €M42 》
21-18=3
35-32=3
42-39=3

D+3 [21,35,42]
21=3×7
35=5×7
42=2×3×7
__________
c.m.m.m.c=2×3×5×7=6×35=210
D+3€M210
D+3={0,210,420...}
D+3=210
D=207
a : 21 = x rest 18 => a = 21x+18 |+3
a :35 = y rest 32 => a = 35y+32 |+3
a : 42 = z rest 39 => a = 42z+39 |+3

=> a + 3 = 21×(x+1)
=> a + 3 = 35×(y+1)
=> a + 3 = 42×(z+1)

21 = 3 × 7
35 = 5 × 7
42 = 2 × 3 × 7
-----------------------
c.m.m.m.c = 2 × 3 × 5 × 7 = 210

deci,
a + 3 = 210
a = 210 - 3
a = 207 (numarul cautat)

Verificare:

207 : 21 = 9 rest 18
207 : 35 = 5 rest 32
207 : 42 = 4 rest 39