Presupun ca abc, bca si cab au bara deasupra, asadar le putem descompune in baza 10 astfel:
abc = 100a + 10b + c
bca = 100b + 10c + a
cab = 100c + 10a + b
s = abc + bca + cab ==> s = (100a+10b+c)+(100b+10c+a)+(100c+10a+b)
s = (100a + 10a + a) + (100b + 10b + b) + (100c + 10c + c)
s = 111a +111b + 111c = 111(a + b + c)
Dar 111 = 3 * 37 ==> s = 3 * 37(a + b + c) ==> s este divizibil cu 3
