tag:blogger.com,1999:blog-15628310.post114173786504352532..comments2022-05-11T09:54:19.451-04:00Comments on Question of the day: Gossip heard round the worldAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-15628310.post-1141823231439574352006-03-08T08:07:00.000-05:002006-03-08T08:07:00.000-05:00The minimum number of calss required is 8. The for...The minimum number of calss required is 8. The formula for such a problem is 2n-4 where n is the number of busybodies.<BR/><BR/>But for this problem, let's label each of the callers as A, B, C, D, E and F. Divide the six into two groups so that A, B, and C are in group 1. Then start calling:<BR/>1) A-B<BR/>2) A-C (A and C now know all there is in group 1)<BR/>3) D-E<BR/>4) D-F (D and F now know all there is in group 2)<BR/>5) A-D (A and D now know everything)<BR/>6) C-F (C and F now know everything)<BR/>7) B-A (B knows everything)<BR/>8) E-F (E knows everything)Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.com