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Ghost Post
Icarian Member

 Posted: Tue Feb 27, 2001 4:02 am    Post subject: 1 Apparently a little miscalculation has been done, or I haven't been paying enough attention in school.
jesternl
Yankee Doodle Dutchie

 Posted: Tue Feb 27, 2001 3:00 pm    Post subject: 2 Yep...kinda makes this puzzle hard to solve... ------------------ GRTZ, Jester www.geocities.com/jesternl/antibill.html The most exciting phrase to hear in science, the one that heralds new discoveries, is not "Eureka!" (I found it!) but "That's funny..." -- Isaac Asimov
Ghost Post
Icarian Member

 Posted: Tue Feb 27, 2001 5:16 pm    Post subject: 3 It doesn't matter whether there are 6 or 7 vaults. Take 1, 2, 4, 8, 16, 32 (and 64 if 7 vaults) bars of each vault and weigh them all together. The total weight will uniquely identify which vaults have 9 kg bars and which ones have 10 kg bars. Assuming there are 7 vaults and 3 of them have counterfeits, here is the list of total weights you will get depending on which are counterfeits. Vault #s which have counterfeits:Total weight 1,2,3:1263 1,2,4:1259 1,2,5:1251 1,2,6:1235 1,2,7:1203 1,3,4:1257 1,3,5:1249 1,3,6:1233 1,3,7:1201 1,4,5:1245 1,4,6:1229 1,4,7:1197 1,5,6:1221 1,5,7:1189 1,6,7:1173 2,3,4:1256 2,3,5:1248 2,3,6:1232 2,3,7:1200 2,4,5:1244 2,4,6:1228 2,4,7:1196 2,5,6:1220 2,5,7:1188 2,6,7:1172 3,4,5:1242 3,4,6:1226 3,4,7:1194 3,5,6:1218 3,5,7:1186 3,6,7:1170 4,5,6:1214 4,5,7:1182 4,6,7:1166 5,6,7:1158 I think this method should work even if the number of vaults with counterfeit bars is unknown. May be when the number is known you can do with fewer bars, I haven't tried. [This message has been edited by kachar (edited 02-27-2001).]
QuikSand
Daedalian Member

 Posted: Thu Mar 01, 2001 2:29 pm    Post subject: 4 If you know exactly how many of the vaults contain fakes, then you get even more leeway to use fewer bricks (than just using the powers of two). You can also eschew taking any bricks at all from the final vault. If you read this puzzle to be 3 fakes from 7 vaults, then you can do the whole thing with only 51 bricks: 1from #1,2 from #2,4 from #3...7...13...24. Any result (# of kgs short of 510) will yield a unique combination of either 2 or 3 vaults that are fakes... if it's 2, you may infer that vault #7 also contains fakes. Amidst all the discussion (in the longer thread) about 3+3=7 and other such things, this is, I believe, the lowest number of bricks that has been offered as a solution to the most logical interpretation of the puzzle (IMO).
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