Question

suma a 4 nr consecutive impare este 416. sa se afle nr

Answer 1

a   /------------/
b  /------------/-2-/
c  /------------/-2-/-2-/
d /------------/-2-/-2-/-2-/
416-12=404
404/4=101
a=101
b=101+2
b=103
c=101+4
c=105
d=101+6
d=107
101+103+105+107=416

Answer 2

a+b+2+c+4+d+6=416
4a+12=416
4a=416-12
4a=404
a=404:4
a=101(primul nr)
b=101+2
b=103 (al doilea nr)
c=101+4
c=105 (al treilea nr)
d=101+6
d=107 (al patrulea nr )
nr sunt:101,103,105,107 
.sper ca te am ajutat .si sa primesc funda