b < a <c < d < 7 a+b+c+d = 18 18 ≥ a+b+c >11 ptr. d=6 ⇒ a+b+c =12 si deoarece c<d ⇒ c=5 a=4 b=3
2,4,6,......58 ....sa le luam la rand : I. 9,6,8,10,...58 II. 18,8,10..58 III. 29,10,12.....58 etc; pana la urma se aduna toate numerele + 3×nr.lor ⇔2+4+6+8+......+58+3·29 =2(1+2+3+...+29) +3·29=29·30 + 3·29 = 29·33=957