Răspuns :
I concurent /-----/
II concurent /-----/----/----/ +4 } 706
III concurent /----/----/-----/-+4/+12
706 - ( 4+4+12 ) = 706 - 20 = 686
1+3+3 = 7 segmente egale
686 : 7 = 98
I concurent = 98 dam
II concurent = 98 x 3 + 4 = 298 dam
III concurent = 298 + 12 = 310 dam
II concurent /-----/----/----/ +4 } 706
III concurent /----/----/-----/-+4/+12
706 - ( 4+4+12 ) = 706 - 20 = 686
1+3+3 = 7 segmente egale
686 : 7 = 98
I concurent = 98 dam
II concurent = 98 x 3 + 4 = 298 dam
III concurent = 298 + 12 = 310 dam
a+b+c=706(dam)
b=3a+4
c=b+12=3a+4+12=3a+16
a+3a+4+3a+16=706
7a+20=706
7a=706-20
a=686/7
a=98(dam)
b=3*98+4=294+4=298(dam)
c=b+12=298+12=310(dam)
R:Primul concurent parcurge o distanta de 98 dam,al doilea de 298 dam iar al treilea de 310 dam
b=3a+4
c=b+12=3a+4+12=3a+16
a+3a+4+3a+16=706
7a+20=706
7a=706-20
a=686/7
a=98(dam)
b=3*98+4=294+4=298(dam)
c=b+12=298+12=310(dam)
R:Primul concurent parcurge o distanta de 98 dam,al doilea de 298 dam iar al treilea de 310 dam