Open Posted December 31, 2014 Report Share Posted December 31, 2014 @Reckon, tu ne dai probleme de matematica, ori te lauzi ca vorbesti cu io.kent? Link to comment Share on other sites More sharing options...
Byte-ul Posted December 31, 2014 Report Share Posted December 31, 2014 Voi sunte?i spar?i? Orice 2 numere impare adunate dau un num?r par. Deci ?i orice 4 numere impare adunate dau tot un num?r par.E imposibil. Link to comment Share on other sites More sharing options...
Faciubici Posted December 31, 2014 Report Share Posted December 31, 2014 Voi sunte?i spar?i? Orice 2 numere impare adunate dau un num?r par. Deci ?i orice 4 numere impare adunate dau tot un num?r par.E imposibil.Nu si daca e o gluma rurala .A zis mai sus ca " nu am zis cati cai cate iepe" Link to comment Share on other sites More sharing options...
io.kent Posted December 31, 2014 Report Share Posted December 31, 2014 Nu este o gluma..Primul grajd 1 Iapa, care naste in grajd !al doilea- 9 Cai..al treilea 9 Cai..si ultimul 5 cai..Total in 4 grajduri 25 de Cai Enjoy happy new year To All of you and your families...have a great one!!! Link to comment Share on other sites More sharing options...
z4rk Posted December 31, 2014 Report Share Posted December 31, 2014 (edited) Fie (2a+1), (2b+1), (2c+1), (2d+1) nr. impare reprezentand nr. de cai din fiecare grajd Avem : (2a+1) + (2b+1) + (2c+1) + (2d+1) = 25, sau 2(a+b+c+d) + 4 = 25, sau 2(a+b+c+d) = 21, ceea ce e imposibil, membrul stang fiind par iar membrul drept impar. Edited December 31, 2014 by z4rk Link to comment Share on other sites More sharing options...
TheTime Posted December 31, 2014 Report Share Posted December 31, 2014 Despre inutilitatea acestui subiect: "un nebun arunca o piatra in lac si 10 intelepti se chinuie s-o scoata de acolo". Link to comment Share on other sites More sharing options...
