Question

se considera numerele naturale a, b,c si d direct proportionale cu 5, 4,2,3.Stiind ca bc=4d, determinati multimea divizorilor numarului a+b+c+d.

Answer 1

{a,b,c,d} d.p. {5,4,2,3}
=> a/5=b/4=c/2=d/3=(a+b+c+d)/(5+4+2+3)=(a+b+c+d)/14=k
=> a=5k
=> b=4k
=> c=2k
=> d=3k
=> a+b+c+d=14k

bc=4d => 4k·2k=4·3k |:(4k) => 2k=3 => k=3/2

a+b+c+d=14k=14 · 3/2=7·3=21

D21={1; 3; 7; 21}
=> a+b+c+d are 4 divizori

(Daca te referi si la divizorii intregi, atunci are 8: -21, -7, -3, -1, 1, 3, 7, 21)