hopeso009 Posted April 21, 2009 Posted April 21, 2009 (edited) Here comes the story_____________________________________________________________ An old man... ordered 10 golden bricks When he receives the bricks, he learns that one of the bricks is not correct in weight Although they look the same in appearance, the special gold brick seems to be either heavier or lighter in weight. Can you kindly help the old man to find the special gold brick by using the balance not more than 3 times? . Edited August 1, 2010 by hopeso009
Raven Brickroft Posted April 21, 2009 Posted April 21, 2009 I already got this in Professor Layton's game for the Nintendo DS. Lemme think this one through. Put 5 / 5 on the scale... no no no... won't work....
keetong Posted April 21, 2009 Posted April 21, 2009 You set it up as 3, 3, 3 and 1 you make the first 3 verse the second 3, if its uneven you take one set and make it verse the third 3, if it goes uneven again you take the one that been either going down or up the whole time and make 1 verse another 1. If its even take the one set of three you didn't weigh for the second time, from the uneven set if it goes uneven take the one which follows the same pattern as the other 2 weigh ins, if its equeal take the one thats from the eneven set that you've havn't used, and that is your brick. If the first 3, second 3 and third three are all equeal its the one that hasn't been touched. I know its a bit hard to understand but its also a bit hard to write its easier in practical.
jifel Posted April 21, 2009 Posted April 21, 2009 You set it up as 3, 3, 3 and 1 you make the first 3 verse the second 3, if its uneven you take one set and make it verse the third 3, if it goes uneven again you take the one that been either going down or up the whole time and make 1 verse another 1. If its even take the one set of three you didn't weigh for the second time, from the uneven set if it goes uneven take the one which follows the same pattern as the other 2 weigh ins, if its equeal take the one thats from the eneven set that you've havn't used, and that is your brick. If the first 3, second 3 and third three are all equeal its the one that hasn't been touched.I know its a bit hard to understand but its also a bit hard to write its easier in practical. Meh. I think I understand what you are saying, and it is correct. Still, bad grammar makes it "???????" MY SOLUTION: Split the ten into 3, 3, 3, and 1. Put two sets of three on the balance. If the two sets of three are uneven, take the first set of three against the third set of three. -If it is uneven, then the first set contains the uneven brick. (two measures are used) Take two of the first set of three, and measure. If they are even it is the last of the set of three. (If they are uneven, then it is one of the two, but which?) -If they are even, then it is the second set that contains the uneven brick. Take two of the second set of three, and measure. If they are even it is the last of the set of three. (If they are uneven, then it is one of the two, but which?) If the two sets of ten are even, take the first set of three against the third set. -If those are even, it is the tenth, lone brick. If those are uneven, it is one in the third set. Take two of the third set of three, and measure. If they are even it is the last of the set of three. (If they are uneven, then it is one of the two, but which?) So with 3 measures, you can narrow it down to two bricks. Humph. jifel
Etzel Posted April 21, 2009 Posted April 21, 2009 Take 5 bricks on each side of the balance. Then take the group that weight less and put 2 bricks on each side. If they are even the fifth brick is the fake one. If they are uneven you just take the two that was less and compare them on the balance. That's 3 times.
Algernon Posted April 21, 2009 Posted April 21, 2009 Take 5 bricks on each side of the balance. Then take the group that weight less and put 2 bricks on each side. If they are even the fifth brick is the fake one. If they are uneven you just take the two that was less and compare them on the balance. That's 3 times. Except the odd brick could be either heavier or lighter in weight. You may just wind up weighing 5 normal gold bricks by choosing the light side. This is a tricky one... I'll have to think it over.
Zepher Posted April 21, 2009 Posted April 21, 2009 He picks them each up. There, using the scale zero times!
Are92 RoWar Posted April 21, 2009 Posted April 21, 2009 The way real prospectors do it: They melt it. Which ever one doesn't melt is the fake. Or just try to balance all of them: 5 on 5. Which ever one is heavier, remove one gold from the less heavier one, and take one off from the heavier one and replace it. Take another one, and repeat.
keetong Posted April 22, 2009 Posted April 22, 2009 Meh. I think I understand what you are saying, and it is correct. Still, bad grammar makes it "???????" Yeah I got confused writing it, and my grammer went right out the door.
hopeso009 Posted April 22, 2009 Author Posted April 22, 2009 OK the answer is Let's number the brick as 1,2,3,4,5,6,7,8,9,10 1st balance : 123 VS 456 2nd balance : 123 VS 789 ------------------------------ if 123 > 456 123 = 789 then the special is among 456 and being lighter in weight 3rd balance : 4 VS 5 if 4=5 then 6 is the one if 4>5 then 5 is the one and vice versa. if 123 = 456 123 > 789 789 have problem if 123 > 456 123 > 789 then 123 have problem if so on... That is the answer
Raven Brickroft Posted April 22, 2009 Posted April 22, 2009 I thought it was heavier or lighter? If 123 =/= 456 How would you know if 123 or 456 bears the special brick? It could be 123 > 456 because 123 has a heavier brick OR 456 has a lighter brick so following this: 1st balance : 123 VS 456 2nd balance : 123 VS 789 ------------------------------- 3rd balance: 456 4 = 5 it does not hold true that 4 > 6 or 4 < 6 because bricks 123 might have the heavier brick. ------------------ Hope someone understands what I am trying to say here.
jifel Posted April 22, 2009 Posted April 22, 2009 (edited) if 4>5 then 5 is the one and vice versa. I am sorry, that doesn't work. If the odd brick out is Heavier OR lighter, than it could be four. Oh, wait... If the 123 is heavier than 456 and =789, than the odd brick must be in 456, AND IS LIGHTER THAN A NORMAL! That's what everyone is missing. so mesure 4 and 5 and the lighter brick is it! I WIN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! (although Hopeso009 does too) jifel Edited April 22, 2009 by jifel
Etzel Posted April 22, 2009 Posted April 22, 2009 Aah, I get it now! That was a tricky (and good!) one.
hopeso009 Posted April 22, 2009 Author Posted April 22, 2009 (edited) I thought it was heavier or lighter?If 123 =/= 456 How would you know if 123 or 456 bears the special brick? It could be 123 > 456 because 123 has a heavier brick OR 456 has a lighter brick so following this: 1st balance : 123 VS 456 2nd balance : 123 VS 789 ------------------------------- 3rd balance: 456 4 = 5 it does not hold true that 4 > 6 or 4 < 6 because bricks 123 might have the heavier brick. ------------------ Hope someone understands what I am trying to say here. I understand and I want to point out 1st balance : 123 VS 456 2nd balance : 123 VS 789 with 1st and 2nd balance, the results have already told you the special brick is heavier or lighter. (except the special brick is no.10) because there are different scenairo if 123 > 456, 123 = 789 (I think you know why 123789 are normal, thus 456 the abnormal group is lighter) if 123 < 456, 123 = 789 (I think you know why 123789 are normal, thus 456 the abnormal group is heavier) if 123 = 456, 123 > 789 (I think you know why 123456 are normal, thus 789 the abnormal group is lighter) if 123 = 456, 123 < 789 (I think you know why 123456 are normal, thus 789 the abnormal group is heavier) Either way answers you heavier or lighter. (hope you understand what I said) Edited April 22, 2009 by hopeso009
Raven Brickroft Posted April 22, 2009 Posted April 22, 2009 Ohhhh~ riiiiiiiight~ Sorry~ Wasn't thinking that one through.
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