Question

Demonstrati ca abc in baza zece+bca in baza zece+cab in baza zece se divide cu 37

Answer 1

   
[tex]\overline{abc} +\overline{bca} +\overline{cab} = \\ =100a+10b+c + 100b+10c+a+100c+10a+b= \\ = 100a+ 100b+100c+10a+10b+10c+a+b+c= \\ =100(a+b+c) +10(a+b+c) + (a+b+c) = \\ = (a+b+c)(100+10+1) = 111(a+b+c) = 37 \times 3(a+b+c)~ \vdots~ 37 \\ \\ \\ \\ \\ \\ a[/tex]