A=[tex] 2^{a} * 3^{b} [/tex]
|D(A)|=numarul divizorilor lui A
Avem urmatoarea formula:
|D(A)|=(a+1)(b+1)
2*A=[tex] 2^{a+1} * 3^{b} [/tex]
|D(2*A)|=(a+1+1)(b+1)=(a+2)(b+1)=3+|D(A)|, deci
(a+2)(b+1)=3+(a+1)(b+1) (rel 1)
3*A=[tex] 2^{a} * 3^{b+1} [/tex]
|D(2*A)|=(a+1)(b+1+1)=(a+1)(b+2)=4+|D(A)|, deci
(a+1)(b+2)=4+(a+1)(b+1) (rel 2)
Din (rel 1) rezulta ca:
(a+2)(b+1)=3+(a+1)(b+1)
(b+1)[(a+2)-(a+1)]=3
b+1=3
b=2
Din (rel 2) rezulta ca:
(a+1)(b+2)=3+(a+1)(b+1)
(a+1)[(b+2)-(b+1)]=4
a+1=4
a=3
Deci A=[tex] 2^{3} * 3^{2} [/tex]=8*9=72
