Open Posted December 31, 2014 Report Posted December 31, 2014 @Reckon, tu ne dai probleme de matematica, ori te lauzi ca vorbesti cu io.kent?
Byte-ul Posted December 31, 2014 Report 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.
Faciubici Posted December 31, 2014 Report 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"
io.kent Posted December 31, 2014 Report 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!!!
z4rk Posted December 31, 2014 Report 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
TheTime Posted December 31, 2014 Report 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".
