Question

determinati toate numerele scrise pe baza 10 de forma ab bara deasupra care sunt cu 36 mai mari decat ba bara deasupra

Answer 1

ab=10a+b
ba=10b+a
10a+b=10b+a+36
10a-a-10b-b=36
9a-9b=36
36:9=4
a-b=4
a ∈ (4,5,6,7,8,9)
b ∈ (0,1,2,3,4,5)
ab ∈ (40,51,62,73,84,95)

Answer 2

ab=36+ba;
⇒ab=10a+b;
⇒ba=10b+a;
⇒10a+b=36+10b+a;
⇒36=[10a+b]-[10b+a]
⇒36=9a+[-9b]
⇒36=9a-9b⇒9[a-b]=36⇒a-b=4⇒a,b∈N⇒a,b-cifre naturale;
⇒a,b∈[[4,0],[5,1],[6,2],[7,3],[8,4],[9,5];
⇒numerele gasite sunt:40,51,62,73,84,95⇒6 numere care indeplinesc conditiile problemei;