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Determinati perechile de numere naturale a si b,stiind ca (a;b) =36 si a+b=324
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Răspuns :

Din (a,b)=36 ⇒36Ia si 36Ib⇒a=36k,b=36l, unde (k,l)=1
Atunci 36k+36l=324 I :36 ⇔k+l=9 ,k,l ∈N
(k,l)={(1,8),(8,1),(2,7),(7,2),(4,5),(5,4)}
(a,b)∈{(36,288),(288,36),(72,252),(252,72),(144,180),(180,144)}