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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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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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