tag:blogger.com,1999:blog-2893315201359920156.post8373081911189707047..comments2023-10-28T20:35:12.856+05:30Comments on Best Brain Teasers: Logic Puzzle Questionlaveshhttp://www.blogger.com/profile/03731980294342074301noreply@blogger.comBlogger14125tag:blogger.com,1999:blog-2893315201359920156.post-2133317918701391162014-11-21T00:06:40.551+05:302014-11-21T00:06:40.551+05:30Try 4 pairs and it fails to light.
Then exchange b...Try 4 pairs and it fails to light.<br />Then exchange batteries from 2 pairs<br />One will work and other wont.<br />Then exchange batteries from other 2 pairs.<br />U get in 8 attemptsAnonymoushttps://www.blogger.com/profile/15901634457482444804noreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-27923289935227484602014-11-11T07:52:00.549+05:302014-11-11T07:52:00.549+05:30Dividing 3+3+2 is fine.
And if U start trying with...Dividing 3+3+2 is fine.<br />And if U start trying with the group of 2 batteries first, only 6 trials are enough. It is also the worst case analysis.As Usual, Unusual..https://www.blogger.com/profile/04275384413166834921noreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-28128911034014643922014-11-11T07:50:50.537+05:302014-11-11T07:50:50.537+05:30Dividing 3+3+2 is fine.
And if U start trying with...Dividing 3+3+2 is fine.<br />And if U start trying with the group of 2 batteries first, only 6 trials are enough. It is also the worst case analysis.As Usual, Unusual..https://www.blogger.com/profile/04275384413166834921noreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-91442594841685630052014-11-11T00:18:18.365+05:302014-11-11T00:18:18.365+05:307
You must first divide the batteries in three gro...7<br />You must first divide the batteries in three groups: a group of two batteries and two groups of three batteries each. Now you have made sure that one of the group has two working batteries. <br /><br />Now both the groups with three batteries can form three possible combinations and the group with two batteries has just one combination. <br /><br />3 + 3 + 1 = 7.Adminnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-33209168138429713352014-11-10T23:09:31.801+05:302014-11-10T23:09:31.801+05:30Absolutely right.
- S KumarAbsolutely right. <br /><br /><br />- S KumarAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-65449459049305492582014-11-10T21:33:04.012+05:302014-11-10T21:33:04.012+05:30brilliant!!brilliant!!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-77791969329307736032014-11-10T21:31:09.057+05:302014-11-10T21:31:09.057+05:30it is 8.....make four pairs and test...in worst ca...it is 8.....make four pairs and test...in worst case scenario all pair will have 1 faulty........take any two pairs...each will have one fault......at max on 4 run you can tell which two are good one.........Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-189731509769643492014-11-10T20:35:10.011+05:302014-11-10T20:35:10.011+05:30to GUARANTEE, 19to GUARANTEE, 19Anonymoushttps://www.blogger.com/profile/09522769958658904190noreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-29107779585597059642014-11-10T17:49:13.021+05:302014-11-10T17:49:13.021+05:30least number is 1 (you were lucky and got them to ...least number is 1 (you were lucky and got them to work without changing)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-50814939593930569542014-11-10T14:02:09.230+05:302014-11-10T14:02:09.230+05:302424Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-2407014562952870162014-11-10T12:41:42.368+05:302014-11-10T12:41:42.368+05:30To get the flashlight on ,one needs 2 good batteri...To get the flashlight on ,one needs 2 good batteries.<br />The optimal way (less trials) is to divide randomly the 8 bat. in 2 groups of 3 ,and 1 group of 2 batteries. <br />Then, according to a famous and very useful mathematical principle called "Pigeonhole Principle", one of the three groups will always have 2 good batteries.<br />Each triplet has c(3,2)=3!/(1!*2!)=3 combinations<br />If the two good ones are in the only couple= 1 combination.<br />So, maximal no. of tests:<br />3+3+1=7 (if we are lucky,we will do it in less tests)RIZOPOULOS GEORGIOShttps://www.blogger.com/profile/05401576457945165575noreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-10606488386299791002014-11-10T09:46:25.156+05:302014-11-10T09:46:25.156+05:3099Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-83387682531582209672014-11-10T09:11:32.576+05:302014-11-10T09:11:32.576+05:3011Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-70430918917247458692014-11-10T09:08:05.476+05:302014-11-10T09:08:05.476+05:301414Anonymousnoreply@blogger.com