Question

Aratati ca nr B=5xy+x3y+xy7 este divizibil cu 3, oricare ar fi cifrele x si y, x diferit de 0

Answer 1

B= 5xy +x3y +xy7 = 500 +10x +y + 100x +30 +y +100x +10y +7 = 210x+12y+537 = 3 ( 70x +4y +179)  ⇒ divizibil cu 3

Answer 2

$B= \overline{5xy} + \overline{x3y} + \overline{xy7}= \\ \\ ~~~=(500+10x+y)+(100x+30+y)+(100x+10y+7)= \\ \\ ~~~=537+210x+12y= \\ \\ ~~~=3 \cdot 179 + 3 \cdot 70x+3 \cdot 4y= \\ \\ ~~~=3(179+70x+4y) ~ \vdots~ 3.$