Răspuns :
abc+bca+cab=100a+10b+c+100b+10c+a+100c+10a+b=111a+111b+111c=111(a+b+c)=37*3(a+b+c) => numerele de forma abc+bca+cab sunt divizibile cu 37.
[tex]abc+bca+cab= \\\\ = 100a+10b+c+100b+10c+a+100c+10a+b= \\\\ = 111a+111b+111c= \\\\ =37(3a+3b+3c) : 37 \ \ \ \ (Trebuiau \ 3 \ puncte)[/tex]