Notam nr cu a,b si c
a+b+c = 1 015-676 = 339
a |____|
b |____|+2
c |____|+4
avem 3 segmente egale +2+4 = 339
339 - 6 = 333
333 : 3 = 111 ( un segment)
a = 111
b = 111+2 = 113
c = 111+4 = 115
Verificare:
111+113+115 = 339
676+339 = 1 015
Numerele consecutive impare sunt:
111; 113; 115.
