a. ab-ba=a·b-a ⇔
10a+b-(10b+a)=a·b-a ⇔
10a+b-10b-a=a·b-a ⇔9a-9b=a·b-a ⇔
9a-9b-a·b+a=0 ⇔
10a-a·b=9b ⇒a(10-b)=9b ⇒adevarat oricare ar fi ab∈N(in baza 10 unde a;b≠0);
b. a(10-b)=9b
9/9 ⇒9/9b ⇔9/ a(10-b)
Avem urmatoarele cazuri:
i. 9/a ⇒a∈{0;9} dar deoarece a≠0 ⇒a=9;
a=9 ⇒9(10-b)=9b ⇒10-b=b ⇒b=5;
Asadar ab=95.
ii. 9/10-b ⇒10-b∈{0;9} dar deoarece b=cifra naturala(in baza 10) ⇒10-b=9 ⇒b=1;
a·9=9 ⇒a=1;
Asadar ab=11;
iii. 3/a ⇒a∈{0;3;6;9} ⇒convine a∈{3;6;9};
3/10-b ⇒10-b∈{0;3;6;9} ⇒convine 10-b∈{3;6;9} ⇔b∈{7;4;1}
a=3 ⇒3(10-b)=9b ⇒30=12b ⇒nu conivine deoarece 12 nu divide 30;
a=6⇒6(10-b)=9b ⇒60=15b ⇒b=4;
Asadar ab=64;
a=9⇒9(10-b)=9b ⇒10-b=b ⇒b=5;
In concluzie ab∈{11;64;95}.