Question

Impartind numerele 563,761 si 441 la acelasi numar natural, obtinem resturile 11,17, respectiv 9. Determinati numerele.

Answer 1

563:n=c1 rest 11  <=> 563=n·c1+11=> n·c1=552

761:n=c2 rest 17 <=> 761=n·c2+17 => n·c2=744

441:n=c3 rest 9  <=> 441=n·c3+9 =>  n·c3=432

n>17, impartitorul>restul

=> n=(552, 744, 432)=24

n=24

Answer 2

563, 761  si 441

D = I ×C + R

563 = I × C₁ + 11 ⇒  I × C₁ = 563 - 11 ⇒I × C₁ = 552

761 =  I × C₂ + 17⇒ I × C₂ =  761 -17⇒ I × C₂ =  744

441 =  I × C₃ + 9⇒  I × C₃ = 441 - 9⇒  I × C₃ = 432

I × C₁ = 552 = 2³ × 3× 23 = 24 × 23

I × C₂ = 744 = 2³ × 3 × 31 = 24  × 31

I × C₃ = 441 = 2² × 2² × 2 ×3² = 24 × 18

deci  impartitorul este 24