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St
post 12 Jun 2005, 20:31
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QUOTE(woth @ 12 Jun 2005, 20:26)
O gaina si jumatate face un ou si jumatate intr-o zi si jumatate.
Cate oua fac 9 gaini in 9 zile?
*



81 oua?
Se face cu regula de 3 compusa,. sar nu o mai stiu..


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post 12 Jun 2005, 20:39
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54 sa fie?


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kukuruz
post 12 Jun 2005, 20:43
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QUOTE(Valentine @ 12 Jun 2005, 20:10)
Gresesti. Daca cea falsa este printre cele 12 pe care le cantaresti? smile.gif

pai, io zic ca trebe 3 cantariri ca sa sa stim excat care e aia falsa


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gattoman
post 12 Jun 2005, 21:02
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36 oua smile.gif tongue.gif
se reduce la:
O gaina face un ou intr-o zi.
Cate oua fac 6 gaini in 6 zile?
sper ca nu am gresit. rolleyes.gif

This post has been edited by gattoman: 12 Jun 2005, 21:05


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Valentine
post 12 Jun 2005, 21:04
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QUOTE(rcosmin @ 12 Jun 2005, 21:14)
pai punand 6 monezi pe un taler si 6 monezi pe un taler (daca toate sunt ok) ar trebui ca balanta sa fie egala intre cele 6 monezi nuh? iar daca cea falsa se afla printre ele ne dam seama unde este,dar daca balanta este  la jumatate inseamna ca moneda falsa o am in mana biggrin.gif  corect? biggrin.gif
*


Cum adica iti dai seama unde este? Cazul cel mai simplu este evident cand ai noroc (exclus smile.gif ) si sunt egale. Dar daca nu sunt? Atentie! Nu stii ca moneda falsa este mai usoara sau mai grea!

kukuruz ai dreptate. Explica cum faci smile.gif (fara ne bazam pe noroc sper smile.gif ).


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elro
post 12 Jun 2005, 21:16
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QUOTE(Valentine @ 12 Jun 2005, 21:04)
Cum adica iti dai seama unde este? Cazul cel mai simplu este evident cand ai noroc (exclus smile.gif ) si sunt egale. Dar daca nu sunt? Atentie! Nu stii ca moneda falsa este mai usoara sau mai grea!

kukuruz ai dreptate. Explica cum faci smile.gif (fara ne bazam pe noroc sper smile.gif ).
*



PAI TU AI ZIS:"una este falsa ce are greutatea diferita de toate celelalte 12. "
DECI!!!
EU inca imi sustin ideea tongue.gif

This post has been edited by rcosmin: 12 Jun 2005, 21:18


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TheSaiLor
post 12 Jun 2005, 21:18
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<<3 barbati se duc la un hotel. Camera costa 30$, deci fiecare plateste 10$. Receptionera isi da seama ca de fapt camera costa doar 25$. Il cheama pe hamal si ii da 5$ sa-i duca inapoi barbatilor. Hamalul nu stie cum sa imparta 5$ la 3 oameni, asa ca le da cate 1$, iar el pastreaza 2. Deci fiecare barbat a platit 9$. 9*3=27, hamalul ia 2$, unde e 1$?.. >>

Mi se pare mie sau voi dati raspuns la cu totul alta intrebare? Deci omu nostru zice asa: 30:3 = 10; 10-1 = 9; 9*3=27; 27+2=29 si in nici un caz nu e 30!! Fara zecimale si alte chestii .. unde e 1$????


Ca sa fiu mai exact ... ei in mod sigur au platit 27$ in final. Tot in mod sigur hamalul ramane cu 2$. Receptionera si aia a dat 5 dolari inapoi deci nu avea cum sa ramana la ea 1$.

This post has been edited by TheSaiLor: 12 Jun 2005, 21:24


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kukuruz
post 12 Jun 2005, 21:28
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QUOTE(Valentine @ 12 Jun 2005, 22:04)
kukuruz ai dreptate. Explica cum faci smile.gif (fara ne bazam pe noroc sper smile.gif ).


pai, dupa ce cantaresc 6 si 6 si daca moneda falsa nu e cea ramasa (a 13-a), atunci ea este intre cele 12, iau cele 6 care trag mai greu si le cantaresc 3 si 3 sa vad care e tripleta care trage mai greu, iar la a 3-a cantarire, 1 si 1, daca sunt egale atunci falsul este moneda ramasa, daca nu, atunci falsul este aia mai grea.

voila! smile.gif


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gattoman
post 12 Jun 2005, 21:35
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e pusa problema gresit. SYL a raspuns corect valuri.gif
e cam asa: prima data au dat cate 10$, apoi au primit inapoi cate 1$, deci pana la urma au dat cate 9$ de om.
9$*3=27$=25$(camera) + 2$(hamal)
inteles? cool.gif


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SYL
post 12 Jun 2005, 21:47
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QUOTE(St @ 12 Jun 2005, 21:27)
PAi toti am zis ca 2$ sunt la hamal...
*



In postul nr 5 ai zis ca sunt la receptionera !!
A, ca dupa ce mai multi au zis ca mine, ai zis si tu ca sunt la hamal... tongue.gif


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madalina
post 12 Jun 2005, 21:51
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Nu stim daca moneda falsa este mai grea sau mai usoara.


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elro
post 12 Jun 2005, 22:11
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QUOTE(madalina @ 12 Jun 2005, 21:51)
Nu stim daca moneda falsa este mai grea sau mai usoara.
*




pai daca este mai grea cantareste mai mult,si daca este mai usoara este tocmai invers.......asa ca ce am zis si io mai sus+ce a zis si kukuruz se adevereste tongue.gif


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post 12 Jun 2005, 22:23
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Valentine
post 12 Jun 2005, 22:57
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QUOTE(rcosmin @ 12 Jun 2005, 23:11)
pai daca este mai grea cantareste mai mult,si daca este mai usoara este tocmai invers.......asa ca ce am zis si io mai sus+ce a zis si kukuruz se adevereste tongue.gif
*


Ori faci misto, ori nu ai nici o treaba cu problema smile.gif

Kukuruz nu este corect. Asa cum am spus, si a accentuat madalina, nu stii daca e mai grea sau mai usoara smile.gif


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kukuruz
post 12 Jun 2005, 23:08
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QUOTE(Valentine @ 12 Jun 2005, 23:57)
Kukuruz nu este corect. Asa cum am spus, si a accentuat madalina, nu stii daca e mai grea sau mai usoarap smile.gif

Ha! Asta complica chestiunea. Te pomenesti ca tre' sa folosesti la un moment dat si moneda care stii sigur din prima cantarire ca nu este falsa. Da' acu' e ora prea tarzie ca sa mai cogitez *yawn*


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GeorgeB
post 12 Jun 2005, 23:11
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valentine te bat sa mor io ... am facut un curs intreg pe problema asta laugh.gif adica 3 ore smile.gif

doar ca eu stiam ca sunt 12 monede. ca e ceva legat de numarul 3 smile.gif


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GoGu
post 12 Jun 2005, 23:45
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Mie imi ies 4 masuratori... Hai sa zic cum am facut... poate am inceput bine, poate (foarte probabil) am luat-o prin stufaris:

Impart monezile in 3 grupuri de 4 si una singura: grupurile sa le numim 4-1, 4-2, 4-3 si 1-1 (moneda singura).

Masuratoarea 1: compar 4-1 cu 4-2
a ) sunt egale -> falsul e in 4-3 sau e 1-1
b ) sunt inegale -> falsul e in 4-1 sau 4-2

Masuratoarea 2:
-pt. a ) compar 4-1 (bun) cu 4-3:
a1) sunt egale -> 1-1 e falsul
a2) 4-3 e mai greu sau mai usor -> falsul e in 4-3 si e mai greu respectiv mai usor (!)

-pt. b ) presupun pt. simplificarea expunerii 4-1 mai greu. Inlocuiesc 4-1 cu 4-3 (bun).
b1) dezechilibrul se pastreaza -> falsul e in 4-2 si e mai usor
b2) balanta se echilibreaza -> falsul e in 4-1 si e mai greu

Deci, din 2 masuratori, fie am identificat falsul (1-1) dar e noroc, fie am identificat grupa de 4 ce contine falsul si am aflat daca e mai greu sau mai usor.

De acum, din 4 bucati, stiind ca falsul e mai greu/usor eu personal nu pot afla care e falsul decat din 2 masuratori (e banal cum, nu mai explic). Se poate cumva dintr-o singura masuratoare?

This post has been edited by GoGu: 12 Jun 2005, 23:54


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Valentine
post 13 Jun 2005, 00:49
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QUOTE(GeorgeB @ 13 Jun 2005, 00:11)
valentine te bat sa mor io ... am facut un curs intreg pe problema asta laugh.gif adica 3 ore smile.gif

doar ca eu stiam ca sunt 12 monede. ca e ceva legat de numarul 3 smile.gif
*


Da, pentru n general (mutari) numarul de monede este (3^n-1)/2. Pentru n=3 este asadar 13 biggrin.gif

Kukuruz: tocmai se presupune ca nu poti sa ai noroc, adica nu te bazezi pe noroc. Poti sa ai ghinion sa fie cel mai prost caz la fiecare masuratoare.

@Gogu: Si totusi se poate exact cu 3 biggrin.gif Ai avut o idee ok, dar nu duce decat la 4 masuratori. Nu ai cum sa o optimizezi.


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post 13 Jun 2005, 12:51
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astea sunt probleme de probabilitati biggrin.gif destea cica fac la facultate laugh.gif


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Valentine
post 13 Jun 2005, 18:27
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QUOTE(rcosmin @ 13 Jun 2005, 13:51)
astea sunt probleme de probabilitati biggrin.gif destea cica fac la facultate laugh.gif
*


Nu are absolut nici o legatura cu probabilitatile smile.gif


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St
post 13 Jun 2005, 21:11
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QUOTE(SYL @ 12 Jun 2005, 21:47)
In postul nr 5 ai zis ca sunt la receptionera !!
A, ca dupa ce mai multi au zis ca mine, ai zis si tu ca sunt la hamal... tongue.gif
*



Eu am zis la misto atunci, ca imi era prea somn ca sa ma gandesc...

QUOTE(gattoman @ 12 Jun 2005, 21:02)
36 oua  smile.gif  tongue.gif
se reduce la:
O gaina face un ou intr-o zi.
Cate oua fac 6 gaini in 6 zile?
sper ca nu am gresit.  rolleyes.gif
*




Era vorba de 9gaini in 9 zile

This post has been edited by St: 13 Jun 2005, 21:14


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post 13 Jun 2005, 21:14
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george zice biggrin.gif

S(n)=3^0+3^1+3^2+ .... +3^n

3*S(n) monede au nevoie de (n+2) cantariri smile.gif


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post 13 Jun 2005, 22:05
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Pentru cei ce vor solutia: CLICK AICI blush.gif


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gattoman
post 13 Jun 2005, 22:20
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sa vina woth sa spuna daca am gresit sau nu. apoi spun cum am facut daca e corect. 04.gif


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Valentine
post 14 Jun 2005, 00:56
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QUOTE(GoGu @ 13 Jun 2005, 23:05)
Pentru cei ce vor solutia: CLICK AICI  blush.gif
*

Opa, vad ca cineva si-a facut tema biggrin.gif biggrin.gif


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post 14 Jun 2005, 07:43
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QUOTE(woth @ 12 Jun 2005, 20:26)
O gaina si jumatate face un ou si jumatate intr-o zi si jumatate.
Cate oua fac 9 gaini in 9 zile?
*


O gaina si jumatate face un ou si jumatate intr-o zi si jumatate.
9 (1,5x6) gaini fac 9 (1,5x6) oua intr-o zi si jumatate.
9 gaini fac 54 (9x6) oua in 9 (1,5x6) zile.

Deci J.W. a avut dreptate.


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post 14 Jun 2005, 09:12
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ai dreptate b64chr. tre sa-mi trezesc neuronii ca au intrat in vacanta si nici nu s-a terminat sesiunea sad.gif


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post 15 Jan 2006, 12:37
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post 15 Jan 2006, 13:05
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post 15 Jan 2006, 13:36
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Asta e mai degraba o iluzie optica. Caroiajul nu e uniform.

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