About this creation
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?
Correct Answer is:
Divide the bricks in groups of 3 pieces each
e.g.
Group A = brick 1,2,3
Group B = brick 4,5,6
Group C = brick 7,8,9
Group D = brick 10
1st balance : A vs B
2nd balance : A vs C
3rd balance : depends on the results of 1st and 2nd balance. Let me explain in following
There are several sceneraio:
Sceneraio [1]
1st balance : A vs B (A heavier)
2nd balance : A vs C (A = C)
which means the special one is among Group B, and being lighter than normal weight.
3rd balance : 4 vs 5 (if equal, 6 is the special one) or (it is the one being lighter in weight)
Sceneraio [2] (almost the same as [1], only it is heavier)
1st balance : A vs B (B heavier)
2nd balance : A vs C (A = C)
which means the special one is among Group B, and being heavier than normal weight.
3rd balance : 4 vs 5 (if equal, 6 is the special one) or (it is the one being heavier in weight)
Sceneraio [3]
1st balance : A vs B (A = B)
2nd balance : A vs C (A = C)
Brick 10 is the special one.

Hope U understand what I say ^_^
Comments


I like it 

February 12, 2011 
But this one got me. I thought I got it but I was looking on it as if the odd one was lighter. I got the answer, luckily. 


I made it 

May 3, 2009 
Quoting Space Pirate
I: 1,2,3,4 vs 5,6,7,8 (1,2,3,4 heavier)
II: 5,6,1 vs 7,8,2 (if equal 3 or 4 is a heavier one, weight 3 vs 4) (7,8,2 heavier => 5 vs 6 the lighter of them is the light one (if both are equal, 1 is a heavy one because of I. round) (5,6,1 heavier => 7 vs 8 (the lighter one) or if equal 2 is a heavy one ;) it works...believe me ;)
got it~~ ^_^
ok it works ~~



I like it 

May 3, 2009 
I: 1,2,3,4 vs 5,6,7,8 (1,2,3,4 heavier)
II: 5,6,1 vs 7,8,2 (if equal 3 or 4 is a heavier one, weight 3 vs 4) (7,8,2 heavier => 5 vs 6 the lighter of them is the light one (if both are equal, 1 is a heavy one because of I. round) (5,6,1 heavier => 7 vs 8 (the lighter one) or if equal 2 is a heavy one ;) it works...believe me ;) 


I made it 

May 3, 2009 
Quoting Space Pirate
hm...I'm pretty sure my answer works though...^^
may be not...^_^....e.g. 1st round = 1,2,3,4 vs 5,6,7,8 (1,2,3,4 heavier) > 2nd round = 1,5,6 vs 2,3,7 (2,3,7 heavier) > 3rd round = how to get it? .....Now either 2 or 3 may be heavier but either 5 or 6 may be lighter too. After 2 rounds, there are still 4 suspect bricks. It seems not very possible to find it by the last round. ^_^



I like it 

May 3, 2009 
hm...I'm pretty sure my answer works though...^^ 


I made it 

May 2, 2009 
add answer 


I like it 

May 2, 2009 
I: 4 bricks vs 4 bricks (if its equal, take one of those 8 normal weight bricks vs one of the two remaining, that's two times possible, so no problem here) otherwise II: now you have to compare 3 bricks vs 3 bricks (2 of the lighter side + 1 of the heavier side vs 2 of the lighter side + 1 of the heavier side) (if its equal again, compare the two remaining heavier side ones, one of the two has to be the heavier one) if one side is lighter, it gets complicated ^^ : take the two bricks, that were on the lighter side in the first round and are now on the lighter side again! III: The lighter one of them is the light brick. If both stay equal, the heavier one of the first round on the opposite side of those two lighter bricks has to be a special heavy brick. But I wouldn't get it, when I read through my description...it's difficult to explain, but I guess it works ^^ anyway...nice one, that kept me thinking for a long time! :D 

so it's almost the same here, but you start with 8 bricks, 4 on each side. ;)
oh, I overlooked the either heavier or lighter thing...well...then it's quite difficult! ^^ 


I made it 

May 2, 2009 
don't give up too soon ^_^ 


I like it 

May 1, 2009 
I give up. : ) 


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