Question

aratati ca daca n = abc + bca + cab (a,b,c, ≠ 0 ) , atunci n ≡ 37 . ≡ ⇒ divizibil sau se imparte exact .

Answer 1

n=100a+10b+c+100b+10c+a+100c+10a+b
n=111a+111b+111c
n=111[a+b+c] ,unde a,b,c≠0;
⇒37/111⇒111/37=3;
⇒n=M37[a+b+c]⇒n este multiplu de 37;