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BMAD

Balancing weights

Question

A balance and a set of metal weights are given, with no two the same. If any pair of these weights is placed in the left pan of the balance, then it is always possible to counterbalance them with one or several of the remaining weights placed in the right pan. What is the smallest possible number of weights in the set?

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11 answers to this question

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Initial observation: it definitely needs more than 4.  I'll see if I can find a solution for 5.

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Still looking for better, but I think I've done it with 7.

3 4 5 6 7 8 9

4 5 6 7 8 almost works, but I can't counterbalance 5+4.

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for a set of five...

Spoiler

...thinking it is not possible.  Label the five integers in increasing order a,b,c,d,e:
e+d has to equal a+b+c and e+c must equal a+b+d but e+d>e+c and a+b+c<a+b+d

 

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1 hour ago, plainglazed said:

for a set of five...

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...thinking it is not possible.  Label the five integers in increasing order a,b,c,d,e:
e+d has to equal a+b+c and e+c must equal a+b+d but e+d>e+c and a+b+c<a+b+d

 

Agree.

Spoiler

I think you have the solution at 6.
I've been doing this in my head for two days. I was only able to eliminate 5.

 

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On 3/10/2018 at 3:17 AM, plainglazed said:

for a set of five...

  Hide contents

...thinking it is not possible.  Label the five integers in increasing order a,b,c,d,e:
e+d has to equal a+b+c and e+c must equal a+b+d but e+d>e+c and a+b+c<a+b+d

 

This contradicts the solution below.  As your a > b > c > d > e > f > g solution does not follow the condition that 

a + b must equal the sum of the rest

On 3/9/2018 at 6:55 PM, plainglazed said:

another seven

  Hide contents

1,2,3,4,5,6,7

and I think an example of six?

  Hide contents

8,7,6,5,4,2

still working on five...

 

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On 3/9/2018 at 6:55 PM, plainglazed said:

another seven

  Reveal hidden contents

1,2,3,4,5,6,7

and I think an example of six?

  Hide contents

8,7,6,5,4,2

still working on five...

multiples of your six seem to work too

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20 hours ago, BMAD said:

This contradicts the solution below.  As your a > b > c > d > e > f > g solution does not follow the condition that 

a + b must equal the sum of the rest

 

yes, but...

Spoiler

...was referring to the specific case of only five weights to prove a solution cannot exist

 

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4 would be the smallest number.  assign a unit weight of 1 ;2 ;3 ;4; to the 4 weights  place units 1 and 4 on left pan and 2 and 3 on the right pan each weigh 5 and balance

sorry I misunderstood the question in my earlier answer

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43 minutes ago, Donald Cartmill said:

4 would be the smallest number.  assign a unit weight of 1 ;2 ;3 ;4; to the 4 weights  place units 1 and 4 on left pan and 2 and 3 on the right pan each weigh 5 and balance

sorry I misunderstood the question in my earlier answer

but if you place 3,4 on one side it would not be balanced.

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