Reckon Posted December 30, 2014 Report Share Posted December 30, 2014 Discutand prin mesaje cu @io.kent nebunu mi-a pus urmatoarea problema:Ai 25 de cai si 4 gradjuri (sau grajde cum se spune), trebuie sa bagi toti caii astia in cele 4 iar numarul de cai admis intr-un grajd minim este 1 si maxim 21, iar caii trebuie sa fie cu numar impar (1,3,5,7,9 etc).Deci problema ar fi:Suma a patru numere impare sa dea ca rezultat 25.Ganditi-va care ar fi ecuatia. Link to comment Share on other sites More sharing options...
Whai_Nooa Posted December 30, 2014 Report Share Posted December 30, 2014 (edited) 1+3+5+7+9 ?//e tarziu frate sunt obosit:)) Edited December 30, 2014 by Whai_Nooa Link to comment Share on other sites More sharing options...
Reckon Posted December 30, 2014 Author Report Share Posted December 30, 2014 1+3+5+7+9 ?!Deci problema ar fi:Suma a patru numere impare sa dea ca rezultat 25. Link to comment Share on other sites More sharing options...
H3xoR Posted December 30, 2014 Report Share Posted December 30, 2014 (edited) #include <iostream>using namespace std;int main(){ int s[] = { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 }, sum = 25; for (int i = 0; i < 11; i++) for (int j = 0; j < 11; j++) for (int k = 0; k < 11; k++) for (int l = 0; l < 11; l++) if (s[i] + s[j] + s[k] + s[l] == sum) cout << s[i] << " + " << s[j] << " + " << s[k] << " + " << s[l] << " = " << sum << endl; return 0;}Cine crede c? exist? o solu?ie, se în?eal? . Edited December 30, 2014 by H3xoR Link to comment Share on other sites More sharing options...
youjnhuhn2 Posted December 30, 2014 Report Share Posted December 30, 2014 matematic e cam imposibil sa imparti 25 de cai vii si intregi in 4 grajduri Link to comment Share on other sites More sharing options...
Active Members MrGrj Posted December 30, 2014 Active Members Report Share Posted December 30, 2014 (edited) Suma a patru numere impare nu va fi niciodata impara Pleci de la o ipoteza gresita// intuiesc ca problema are o rezolvare 'haioasa'.Matematic nu ai cum sa faci asta. Edited December 30, 2014 by MrGrj Link to comment Share on other sites More sharing options...
Reckon Posted December 30, 2014 Author Report Share Posted December 30, 2014 @io.kent imi spune pe Whatsapp ca are rezolvarea, deci el trebuie sa vina cu ea (maine probabil) Link to comment Share on other sites More sharing options...
andrei98M Posted December 30, 2014 Report Share Posted December 30, 2014 4(2k+1)= 25 , asta e ecuatia, cred Link to comment Share on other sites More sharing options...
xTremeSurfer Posted December 30, 2014 Report Share Posted December 30, 2014 maxim 6 cai frumosi ! Link to comment Share on other sites More sharing options...
Pacalici Posted December 30, 2014 Report Share Posted December 30, 2014 Trebuie sa pui neaparat in fiecare gradj? Nu este mentionat si atunci merge 1+5+19 Link to comment Share on other sites More sharing options...
Active Members MrGrj Posted December 30, 2014 Active Members Report Share Posted December 30, 2014 Trebuie sa pui neaparat in fiecare gradj? Nu este mentionat si atunci merge 1+5+19Ba scrie.trebuie sa bagi toti caii astia in cele 4 Link to comment Share on other sites More sharing options...
Viral-One Posted December 30, 2014 Report Share Posted December 30, 2014 (edited) Matematic, un numar trebuie sa fie obligatoriu PAR ( 5(2+1+1+1) = 10+5+5+5= 25 ) dar poate ma insel, astept solutia lui @io.kent, este interesanta problema, sunt curios daca o sa fie cineva care o sa gaseasca solutia corecta Edited December 30, 2014 by Viral-One Link to comment Share on other sites More sharing options...
Pacalici Posted December 30, 2014 Report Share Posted December 30, 2014 (edited) Sa incapa in cele 4, eu unu nu inteleg din asta ca nu trebuie sa ramana un grajd gol. In fine, o sa astept rezolvarea.De asemeni nu este mentionat ca numerele sa fie diferite si atunci exista alte variante. Edited December 30, 2014 by Pacalici Link to comment Share on other sites More sharing options...
Elohim Posted December 30, 2014 Report Share Posted December 30, 2014 E simplu.In primul il punem pe CIn al doilea il punem pe AIn al treilea in punem pe IIn al patrulea il punem pe IGata, am bagat caii in cele 4 grajduri. Link to comment Share on other sites More sharing options...
Reckon Posted December 30, 2014 Author Report Share Posted December 30, 2014 E simplu.In primul il punem pe CIn al doilea il punem pe AIn al treilea in punem pe IIn al patrulea il punem pe IGata, am bagat caii in cele 4 grajduri.No way, no way Problema zilei: 40 de cai in 9 grajduri Link to comment Share on other sites More sharing options...
sleed Posted December 30, 2014 Report Share Posted December 30, 2014 Rezolvarea mea ar fi asta : Faci produsul mezilor, belesti Pu*a iezilor, trimiti iezii la pascut si iei capra la Pul* daca nu ti da rezultatul o iei de la cap cu tapulfaci produsul mezilor belesti Pu*a iezilor, trimiti iezii la pascut si iei capra la Pul* daca nu ti da rezultatul o iei de la cap cu tapul,faci produsul mezilor belesti Pul* iezilor trimiti iezii la pascut si iei capra la Pul* daca nu ti da rezultatul o iei de la cap cu tapul Link to comment Share on other sites More sharing options...
Elohim Posted December 30, 2014 Report Share Posted December 30, 2014 No way, no way Problema zilei: 40 de cai in 9 grajduriYes way, asta e singura solutie Link to comment Share on other sites More sharing options...
Pacalici Posted December 30, 2014 Report Share Posted December 30, 2014 Reckon a fost mai rapid, vroiam sa pun si eu link si atunci Elohim a avut dreptate. Link to comment Share on other sites More sharing options...
Reckon Posted December 30, 2014 Author Report Share Posted December 30, 2014 Il asteptam pe @io.kent maine cu rezolvarea, pana atunci, think about it, eu am langa mine vreo 3 pix-uri, cateva liniare, si incerc sa fac ceva )) Link to comment Share on other sites More sharing options...
andrei98M Posted December 30, 2014 Report Share Posted December 30, 2014 4(2k+1)= 25 , asta e ecuatia, cred#include <iostream>using namespace std;int main(){ int s[] = { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 }, sum = 25; for (int i = 0; i < 11; i++) for (int j = 0; j < 11; j++) for (int k = 0; k < 11; k++) for (int l = 0; l < 11; l++) if (s[i] + s[j] + s[k] + s[l] == sum) cout << s[i] << " + " << s[j] << " + " << s[k] << " + " << s[l] << " = " << sum << endl; return 0;}Cine crede c? exist? o solu?ie, se în?eal? .4(2k+1)= 25 , asta e ecuatia, crednumai daca faci 25/4 iti dadeam seama Link to comment Share on other sites More sharing options...
io.kent Posted December 31, 2014 Report Share Posted December 31, 2014 Deci va mai las sa va mai ganditi dupa aia va dau solutia... Link to comment Share on other sites More sharing options...
andrei98M Posted December 31, 2014 Report Share Posted December 31, 2014 Deci va mai las sa va mai ganditi dupa aia va dau solutia...da macar pe pm sa nu mor prost=)) Link to comment Share on other sites More sharing options...
io.kent Posted December 31, 2014 Report Share Posted December 31, 2014 Va dau o pista, nu am specificat cati cai si cate iepe! Link to comment Share on other sites More sharing options...
devilox Posted December 31, 2014 Report Share Posted December 31, 2014 numarul de cai admis intr-un grajd minim este 1 si maxim 2121 in primul grajd si cate 1 in restu .. 3 cai vor sta like a boss in penthouse se trag la sorti seara inainte de somn cine sta singur Link to comment Share on other sites More sharing options...
JIHAD Posted December 31, 2014 Report Share Posted December 31, 2014 gata ma, a zis gluma. nu va mai obositi Link to comment Share on other sites More sharing options...