Question

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

Answer 1

direct proportionale⇒a/5=b/4=c/2=d/3=k
a=5k,b=4k,c=2k,d=3k
a+b+c+d =5k+4k+2k+3k=14k
inlocuind in bc=4d valorile in functie de k⇒4k.2k=4.3k⇒k=3/2 
dar a+b+c+d=14k=14.3/2=7.3=21
D21={1,3,7,21}