tag:blogger.com,1999:blog-8289183275234086319.post8108051910291182124..comments2019-04-01T00:30:43.611-07:00Comments on {inte{r}esting} - ઇન્ટરેસ્ટીંગ: Student@competitionUnknownnoreply@blogger.comBlogger17125tag:blogger.com,1999:blog-8289183275234086319.post-77864592995885139922010-11-16T20:18:35.738-08:002010-11-16T20:18:35.738-08:00"Ans. of student @ competition-17/11/2010 ,sh..."Ans. of student @ competition-17/11/2010 ,shatdal"<br /><br />no. of boxes=3;<br />no. of black marbles=3;<br />no. of white marbles=3;<br /><br />now let us assume that,<br /><br />box a is labeled=white-white<br />box b is labeled=white-black<br />box c is labeled =black-black<br /><br />now as said in problem there is no any black marble in black-black labeled box c.<br /><=> box c must contain both white marbles.-----------(1)<br />so, without opening box c we can change the label to white-white.<br /><br /><br />now for box a and b the solution is as below:<br /><br />both label are wrong for both boxes.<br /><br />for box a:<br />possible labels are:<br /><br />white-black or black-black<br /><br />for box b:<br />possible labels are:<br /><br />white-white or black-black<br /><br />now here for box b we can not label white-white because we have already labeled c with white-white<br />(from eq.no.(1))<br /><br /><=> box b remains with only one possibility of label black-black-------------(2)<br /><br />and finally remaining one label of white-black goes with box a-----(3)<br /><br /><br /><=> from eq.(1),(2),&(3) without opening a single box we are able to change the labels.<br /><br /><br /><br /><br />ans=0 box should be open<br /><br /><br /><br />regards,<br />Hiren Anand Shah<br />B.TECH-IT(2nd year)<br />Dharmsinh Desai University,nadiad <br /><br />25,shankar park so.<br />o/s ahmedabadi gate,<br />nadiad-gujarat-387001<br /><br />mobile:+91 9067997529<br /><br />e-mail:hirenshah111@gmail.comHi!ren $hahhttps://www.blogger.com/profile/07275141049169401218noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-80064499008172982202010-10-26T08:45:05.912-07:002010-10-26T08:45:05.912-07:00Three thieves A,B and C went to the mango garden. ...Three thieves A,B and C went to the mango garden. They thest mangoes. They hid in the temple. At night they slept there. After some time C got up. He saw A and B were sleeping. He counted mangoes. He found that there was one more mango if the manoges were qually divided. Their are mango and his part he hide and slept. After some time B got up. He saw A and C sleeping. He found that there was an more mango if the mangoes were equally divided. This mango and his part he hide and slept. A also had done the same. At the early morning they counted mangoes. They found that there was one more mango of the mangoes were equally divided. They put this mango in the temple and took their own part. Now, question is how many mangoes did they stead.?<br /> Suppose, At the last in distribution they got x mangoes each. so, there were 3x + 1 mangoes. When A get up and went to the heap of mangoes there would be 3x + 1 + 3x + ½ + 1 mangoes.<br /> By simplification it is 9x + 5 /2. When B got up and went to the heap of mangoes there would be 9x + 5 /2 + 9x + 5 /4 +1 mangoes.<br /> By simplification it is 27x +19 /4. when C got up and went to the heap of mangoes there would be.<br /> By simplification it is so they stale mangoes.<br /> 79 is the smaller number sati sting the question. In genral, for every solution. It is a sequence whose first term is 7 and common difference is 8.<br /> x = a + ( n – 1 ) d<br /> = 7 + ( n – 1 )<br /> = 8n – 1 <br /> Put x = 8n – 1 into 81x + 65 /8<br /> 81n – 2 is the genral formulla for the every solutions. All the solutions are 79,160,241,322, 403...Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-55506041360652831622010-10-25T09:59:50.726-07:002010-10-25T09:59:50.726-07:00HI I'M ENGINEERING STUDENT.I WANT TO SALVE DIF...HI I'M ENGINEERING STUDENT.I WANT TO SALVE DIFFICULT PUZZLES. LET THE X=TOTAL પેંડા<br /><br />(1) AT 12:01AM FRIEND1 EAT X પેંડા AND LEAVE K1 પેંડા.<br /><br />SO K1=(X1-1)\3...<1><br /><br />(2) AT 12:11AM FRIEND2 EAT K1 પેંડા AND LEAVE K2 પેંડા<br /><br />SO K2=(K1-1)\3...<2><br /><br />(3) AT 12:21AM FRIEND3 EAT K2 પેંડા AND LEAVE K3 પેંડા<br /><br />SO K3=(K2-1)\3...<3><br /><br />(4) AT સવારે FRIEND1 EAT K3 પેંડા AND LEAVE K4 પેંડા(ત્રણેય મિત્રો પેંડા ખાઈ ગયા)<br /><br />SO K4=(K3-1)\3 =0(ત્રણેય મિત્રો પેંડા ખાઈ ગયા)...<4><br /><br />SOLVING EQ 1,2,3,4 WE GET THE <br /><br />X=TOTAL પેંડા=121<br /><br /><br />THE TOTAL પેંડા 121...ANSGajjar Tejas Rameshbhaihttps://www.blogger.com/profile/13033167108009886328noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-88499195925794753972010-10-24T22:32:23.228-07:002010-10-24T22:32:23.228-07:0079 Penda hata
Falguni Doashi vadodara79 Penda hata<br />Falguni Doashi vadodarafalgunihttps://www.blogger.com/profile/16524619954513566595noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-40826914657369567262010-10-21T10:20:07.574-07:002010-10-21T10:20:07.574-07:00end game 20/10/10
my ans is : 79 pendaend game 20/10/10<br />my ans is : 79 pendaDr hitesh bhaliya (baroda)noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-27542536181078410602010-10-21T08:35:16.954-07:002010-10-21T08:35:16.954-07:00the answer is 79 Sumit Valuthe answer is 79 Sumit Valusumit valuhttps://www.blogger.com/profile/10663222850485962154noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-29776413845599515922010-10-21T03:50:25.428-07:002010-10-21T03:50:25.428-07:00Dear Alpeshbhai,
Considering total no of sweets ar...Dear Alpeshbhai,<br />Considering total no of sweets are X<br />So total sweets eaten by 1st person <br /><br /> ( X-1)/3<br /><br />So total sweets remained after 1st person has eaten his share is <br /><br /> 2*( X-1)/3<br /><br />So total sweets eaten by 2nd person [{(2 X-2)/3}-1]/3<br /> = (2X – 5) /9<br />So total sweets remained after 2nd person has eaten his share is <br /><br /> 2 * (2X – 5) /9<br /><br />So total sweets eaten by 3rd person {(4 X -10)/9 }– 1)/3<br /> = (4 X -19 )/27<br />So total sweets remained after 3rd person has eaten his share is <br /> <br /> 2 * (4 X -19 )/27<br /><br />This is again shared by all the three friends<br /><br />So total sweets for each person in will be <br /> [{(8X -38)/27}/1]/3<br /> <br />= ( 8 X -65)/81 <br /> <br />= (8 X + 16 -81)/81<br /> <br />= ( 8 * (x+ 2)/81) - 1<br /><br />Now no one has broken the sweet so <br /><br />(x + 2)/81 shall be integer. <br /><br />Therefore (X + 2) / 81 = A <br />Therefore X = 81 A -2<br />Now for minimum amount of sweet take A = 1 => X = 79<br />So total no of sweet taken were 79 pieces<br /><br /><br />Dipali V Shah<br />KapadvanjDipalihttps://www.blogger.com/profile/15115490824686268198noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-75116336003517751032010-10-20T23:30:11.468-07:002010-10-20T23:30:11.468-07:00160 penda hata160 penda hataronak kamanihttp://ronakkamani.blogspot.com/noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-56747654499630825242010-10-20T23:06:57.843-07:002010-10-20T23:06:57.843-07:00Iska ans:79
pehla aapne penda ni sankhya dharo ke...Iska ans:79<br /><br />pehla aapne penda ni sankhya dharo ke 3x+1 6e.......<br /><br />have 1st mitra 3 bhag kare 6e ne 1 penda kutra ne nakhe 6e.<br /><br /> 1st 2nd 3rd <br /> x x x 1<br /><br />have baki vadhya penda=2x.<br /><br />have 2nd mitra 3 bhag kare 6e ne 1 pendo kutra ne nakhe 6e<br /><br />have dharo ke 2x=3y+1............ 3y+1 <br /> x = -------- ........................<1<br /> 1st 2nd 3rd 2<br /> y y y 1<br /><br />have baki vadhya penda=2y.<br /><br />have 3rd mitra 3 bhag kare 6e ne 1 pendo kutra ne nakhe 6e<br /><br />have baki vadhya=2y<br /><br />have dharo ke 2y=3z+1........ .3z+1<br /> y = ------- ........................<2 <br /> 1st 2nd 3rd 2<br /> z z z 1<br /><br />have baki vadhya penda=2z.<br /><br />have badha mitro savare penda na 3 bhag kare 6e ne 1 pendo kutra ne nakhe 6e.......<br /><br />have dharo ke 2z=3w+1........... 3w+1<br /> z = -------- ...........................<3<br /> 1st 2nd 3rd 2<br /> w w w 1<br /><br />now from equestion 1,2,3 ..........<br /><br />from equ (3) & (2)<br /><br /><br /> 9w+5<br /> y = ---------- ......................................<4<br /> 4 <br /><br />from equ (4) & (1)<br /><br /><br /><br /> 27w+19<br /> x = ---------- ......................................<5<br /> 8<br /><br /><br />now penda = 3x+1<br /><br /><br /> 81w+19 <br />so 3x+1= -------------<br /> 8 <br /><br />have na-nama nani sankyake je aa badha equestion solve kare ke je badha equestion ma fraction vina ni sankhaya aave te six(6) chhe.<br /><br />so 3x+1=79................final ans...<br /> <br />my name is parth mehta from ahmedabad , vejalpurAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-59423420615123521872010-10-20T22:42:54.713-07:002010-10-20T22:42:54.713-07:00there is good diet for brain to keep it alive
the...there is good diet for brain to keep it alive <br />there were 79 PENDA at starting <br />Thanks <br />jaydip patel<br />E-mail:j_dip@rediffmail.comAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-69316121837081314082010-10-20T21:04:25.222-07:002010-10-20T21:04:25.222-07:0079=26+26+26+1
52=17+17+17+1
34=11+11+11+1
22=7+7+7...79=26+26+26+1<br />52=17+17+17+1<br />34=11+11+11+1<br />22=7+7+7+1<br /><br /><br />Mitesh Patel (Falu-VIjapur)<br /><br />Cont No 9724026672miteshhttps://www.blogger.com/profile/13547607233282332996noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-67769002957838595552010-10-20T20:59:51.023-07:002010-10-20T20:59:51.023-07:0079=26*3+1
52=17*3+1
34=11*3+1
22=7*3+1
Ronak Rava...79=26*3+1<br />52=17*3+1<br />34=11*3+1<br />22=7*3+1<br /><br />Ronak Raval (Deodar)<br />Mob.No 9374023241ronak ravalhttps://www.blogger.com/profile/17278158574117517645noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-68431579918020378202010-10-20T12:19:07.424-07:002010-10-20T12:19:07.424-07:00no of sweetmeat is 73no of sweetmeat is 73jain shraddhanoreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-37308984275320105852010-10-20T08:26:49.396-07:002010-10-20T08:26:49.396-07:0079 pedas... the equation is as follows:
26+26+26+1...79 pedas... the equation is as follows:<br />26+26+26+1=79<br />17+17+17+1=52<br />11+11+11+1=34<br />07+07+07+1=22<br /><br />Dr Samir Dani (Ahmedabad)Dr.Samirhttps://www.blogger.com/profile/03151557012720257922noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-32159349907015272882010-10-20T07:12:34.572-07:002010-10-20T07:12:34.572-07:0079 penda.....Honey P.shah79 penda.....Honey P.shahAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-30770115153637128972010-10-20T06:44:50.886-07:002010-10-20T06:44:50.886-07:00Answer would be 79.
26+26+26+1
factoring 52
17+17+...Answer would be 79.<br />26+26+26+1<br />factoring 52<br />17+17+17+1<br />same way<br />11+11+11+1<br />and the last<br />7+7+7+1.<br /><br />viral fichadiyaviralhttps://www.blogger.com/profile/18340343255482467062noreply@blogger.comtag:blogger.com,1999:blog-8289183275234086319.post-63275757347951695342010-10-20T06:30:33.474-07:002010-10-20T06:30:33.474-07:00very good Alpeshbhai, thanks for giving mental exe...very good Alpeshbhai, thanks for giving mental exercise<br />CA Mukund TrivediAnonymousnoreply@blogger.com