x+y+z = 2013
Daca x-1, y-3 si z-20 sunt numere naturale consecutive in ordinea data, notam x-1 = a , y-3 = a+1 si z-20 = a+2.
Rezulta a + a + 1 + a + 2 = x - 1 + y - 3 + z - 20 <=>
<=> 3*a +3 = x + y + z - 24 <=>
<=> 3*a + 3 = 2013 - 24 = 1989 <=>
<=> 3*a = 1986 => a = 662 => x=663, y=666, z=684
Verificare : 663+666+684 = 2013
