a + b + c = 252
a |_____|
b |_____|+2
c |_____|+4
252 - (2+4) = 252 - 6 = 246
246 : 3 = 82 (un segment egal)
a = 82
b = 82+2 = 84
c = 82+4 = 86
verificare:
82+84+86 = 252
cele 3 numere consecutive pare:
82 ; 84 ; 86 .
sau
a -2 + a + a - 4 = 252
3×a = 252 + 2 + 4
3×a = 258
a = 258 : 3
a = 86
a - 2 = 86 - 2 = 84
a - 4 = 86 - 4 = 82
