tag:blogger.com,1999:blog-98134167249891941.post6516613475603877431..comments2023-07-07T04:22:12.300-07:00Comments on Rick Lime's Limericks: Counterfeit CoinsAnonymoushttp://www.blogger.com/profile/09099835477346969687noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-98134167249891941.post-40035253374928920272013-05-06T07:30:43.914-07:002013-05-06T07:30:43.914-07:00Thanks Sue, your solution does seem more straightf...Thanks Sue, your solution does seem more straightforward than Catherine’s. One of the goals of Catherine’s Pastimes is to explain the process Catherine follows to solve each puzzle. This is a very difficult puzzle and I am not happy with her explanation. I will revise it before publication.<br /><br />The difficult part of this problem is the choice and orientation of coins for the second weighing, if the first weighing doesn’t balance. Thanks to google, I am now aware of a 3rd solution, so the 3 options for the second weighing are:<br />Catherine’s (HLLn) --- (LLnn) <br />Yours (HHL ) --- (HHL )<br />Google's (HLLL) --- (Lnnn)<br /><br />After some consideration, I think the 3rd one may be easier for Catherine to explain.<br />Anonymoushttps://www.blogger.com/profile/09099835477346969687noreply@blogger.comtag:blogger.com,1999:blog-98134167249891941.post-61742779272133784752013-05-05T23:02:42.398-07:002013-05-05T23:02:42.398-07:00While Heathcliff was looking for his limerick-writ...While Heathcliff was looking for his limerick-writing scratch pad, Catherine did some more thinking. Her solution had been arrived at after a good deal of trial and error, but what made it work was that, in the second weighing, there were *not* two coins suspected of being heavy in one pan opposite two suspected of being light in the other, in other words not four doubtful coins. <br />What if she had started afresh at the second weighing, set aside the four coins known to be good, and simply placed two suspected of being heavy coins and one suspected of being light in each pan, leaving two suspected of being light on the table. Then, if the scales balanced, one of the two coins on the table was the light one, to be identified in a third weighing. And if the scales didn't balance, the two suspected heavy ones on the lower pan and the one suspected light one on the upper pan were the new suspects.<br />"Heathcliff, I think I've improved on my solution", she ventured, "but you are the smart one. It seems so simple now, but perhaps I'm missing something?" <br /><br />(From Sue) <br />Anonymousnoreply@blogger.com