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Multipan Balance (Posted on 2009-01-18) Difficulty: 3 of 5
A multipan balance scale depicted below has one pan on the left and five on the right.
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The pan on the left is a unit distance from the fulcrum. The pans on the right are 1, 2, 3, 4, and 5 units from the fulcrum. The balance scale is operated by placing an unknown weight on the left and placing some or all of a known set of weights on the pans on the right.

If the known weights are 1, 2, 3, 4 and 5, show that any integral weight 1 to 75 can be measured. What is the smallest measurement requiring 3 weights on the right? 4 weights? 5 weights?

If the restriction that only one weight can occupy a pan is added, show that any integral weight 1 to 55 can be measured. What is the smallest measurement requiring 3 weights on the right? 4 weights? 5 weights?

No Solution Yet Submitted by Brian Smith    
Rating: 3.0000 (1 votes)

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Proof of first part by mathematical induction Comment 3 of 3 |
If the known weights are 1, 2, 3, 4 and 5, show that any integral weight 1 to 75 can be measured.

If weight is 1 on left, we put weight 1 in pan 1 (i.e. 1 unit from fulcrum).

Now supposing some unknown weight 'k' (such that 1<=k<75 and k is integer) is on left and can be measured, it is to be proved that k+1 can be measured too. With k on left and some weights on right to balance it off, we can assume that at least one weight 'w' is NOT on the pan 5 (i.e. either on one of pans 1 to 4 or off the balance).

Also, weight 1 is either on the balance or off the balance. If weight 1 is off the balance then then we can put weight 1 on pan 1 and thus be able to measure weight k+1.

If weight 1 is on the balance then it is either on pan 5 or not on pan 5. If it is not on pan 5 then we can move weight 1 one unit away from fulcrum and thus be able to measure weight k+1.

Finally if weight 1 is on the balance and it is on pan 5. Here we bring the weight 'w' described above into play. If weight w is off balance then we put it on pan and move weight 1 (w - 1) number of pans toward fulcrum. Hence increasing the net weight on right hand side by 1. Therefore we can measure weight k+1.

If weight w is on the balance then it is NOT on pan 5 (that is how we have defined weight w above). In that case we move weight w one unit away from fulcrum and move weight 1 (w - 1) number of units toward fulcrum. Hence increasing the net weight by 1 unit. Therefore we can measure weight k+1.

Therefore whenever k (1<=k<75 and k is integer) can be measured k+1 can be measured as well. Also weight 1 can be measured as well. Thus by mathematical induction, all integral weights from 1 to 75 can be measured.

What is the smallest measurement requiring 3 weights on the right?

That must be 6 units when weights 1, 2 and 3 are on pan 1.

4 weights?

10 units with weights 1, 2, 3 and 4 on pan 1.

5 weights?

15 units with weights 1, 2, 3, 4 and 5 on pan 1.

[on the second part, later...]

  Posted by elementofsurprize on 2009-01-22 19:48:39
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