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Weighing Bells



On each pan of the tri-beam scale are 3 seemingly identical bells that are balanced with the other pan bells.


A) Sort out the bells-3 light bells (L=1) and 3 heavy bells (H=3) from Normal bells (N=2) by using the scale .
How many times will it take to identify their weights?

B) When you got a desired outcome on the first weighing ,how can all the bells be sorted on the 2nd weighing?

The scale works as follows:
1)Balanced : 3 weights on pans are equal ex. 6 : 6 : 6
2)One pan down : Only the heaviest ex. (5):3 :2 , (5):1:1
3)One pan up : 2 heavier weights are equal ex. (3):4 :4


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2 answers to this question

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Posted (edited)

Part A ) I can identify weights of all bells in 6 weighings or less. I don't know what B ) means.

In what follows, "L" means "Low", "H" means "High". The results of a weighing are always Equal (all three pans equal weight), LL (two Low pans are equal, third is lighter), L (one Low pan is lowest, other two pans are lighter and indeterminate)

1: Weigh three x three x three
1a:Equal 321,321,321
1b:LL: 331, 322,112
1c:L:333,221,211; 333,222,111; 332,322,111; 332,123,112; 332,311,122; 
331,222,311; 331,321,221; 322,321,311

Case 1a:

1a2 weigh a 321
1a2 L shows which is 3
1a23 weigh another 321
1a23 L shows which is 3
1a234 weigh remaining 321
1a234 L shows which is 3--you now know all 3s
1a2345 weigh one each from the 21,21,21
1a2345a: Equal: 111 or 222
1a2345a6: weigh two from one and 1 from the other
1a2345a6a: LL These are the 2s, the other is a 1--(6)
1a2345a6b: L This is the 2, the others are 1s--(6)
1a2345b: LL These are 2s, the other is a 1
1a2345b6: weigh the remaining three
1a2345b6: L: this is the remaining 2, the others are 1s--(6)
1a2345c: L This is a 2, the other two are 1s
1a2345c6: weigh the remaining three
1a2345c6: LL: these are the remaining 2s, the other is 1--(6)

Case 1b:

1b2 weigh one of the pans, either 322 or 331 (you don't know which)
1b2a: LL: This is the 331, other is the 322; one more weighing per pan: (5)
1b2b: L: This is the 322, other is the 331; one more weighing per pan: (5)

Case 1c: 

1c2 weigh the Low pan
1c2a Equal: 333,221,211 or 333,222,111;
1c2a3: weigh another pan
1c2a3a Equal: 222 or 111; weigh two of one against one of the other--(4)
1c2a3b LL: 221, weigh the 211 pan to identify them--(4)
1c2a3c L: 211, weigh the 221 pan to identify them--(4)
1c2b LL: 332,322,111 or 332,123,112 or 332,311,122 or 331,222,311 or 331,321,221
1c2b3: weigh one of the other pans
1c2b3a: Equal 332,111,322; 331,222,311; weigh remaining pan to find the 3; weigh two of the middle pan against one of the H of third pan--(5)
1c2b3b: LL: 331,221,321; weigh the 321 pan to find the 3;  weigh the 21 against a known 2 to find the 1;--(5)
1c2b3c: L: 332,322,111 or 332,312,211 or 332,211,312 or 332,311,122 or 331,311,222 
1c2b3c4: weigh the other pan
1c2b3c4a: Equal 332,322,111 or 331,311,222; weigh the High from first weighing with two of the Equal--(5)
1c2b3c4b: LL 332,311,221--(4)
1c2b3c4c: L 332,312,211 or 332,211,312;
1c2b3c4c5: weigh two High from 3rd pan, one High from 2nd pan
1c2b3c4c5a: Equal: all 1s; weigh L from 3rd pan against two 3s to see if it's 2 or 3--(5)
1c2b3c4c5b: L: 2 and 11, you know whether the ones came from 211 or split from 211,312--(5)
1c2c L: 322,321,311; you know the 3,2,2: weigh another pan to find the 3; weigh other pan to find the other 3; weigh any three remaining bells to find the 2--(5)


Edited by CaptainEd
I wanted to embed spoilers to save real estate

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Hi CaptainED

Actually it is a "case1a" as stated from the OP..Can this be done in 4 or less?

I almost forgot what B) ask..its like placing the bells on the scale and wish (granted)

that the would be balance or one pan up or one pan down..so that on the next weighing

all bells are identified.


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