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Ball Series Insertion (Posted on 2016-07-29) Difficulty: 3 of 5
I have a series of 25 balls whose weights are in arithmetic progression. 24 of the balls have already been sorted and put into ascending order.

Using only a balance scale at most three times, devise a strategy which finds where the last ball belongs in the series.

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In the firs use of the balance the strategy should point to restrict from 24 to 8 the number of possible places in the scale of order for the last ball.  

First measure:

This could be achieved labeling the sorted balls in ascendent order from 1 to 24 and placing all the labeled balls in the balance (so that the ball we want to order is out of the balance). 
We place them as follow: 

A    |    B
======
1          3
2        
---- |  ----
    4 
6     5
---  |  ----   
    7
9          8
---  |  ----
10        
11       12
---  |  ----
13          
14       15
---  |  ----
           16 
18       17
---  |  ----
           19
21       20
---  |  ----
22       
23       24
_________

With these numbers the sum of labels for A and B is the same.

Now, what happen if the unsorted ball is the first in the scale? The real weight of each ball will be one unit higher than their number of ball. The balance will be anyway balanced. 

And if the unsorted ball is the second in the scale? The balance will be unbalanced with more weight on the B side of the scale.

And if it is the third? To make it shorter we will get this: 

1,5,7,11,13,17,19,23  Balanced
6,8,9,10,18,20,21,22 More weight on side A
2,3,4,12,14,15,16,24 More weight on side B

So we have achieved our goal for the first use. We can take the eight balls labeled pointet by the first scale's measure and try to sort them in groups of 3-2-3 balls. 

Second measure:

If the first measure we got was "balanced" we take 1,5,7,11,13,17,19,23 
and put them again in the balance with this disposition: 
A: 1,5,19,23        B: 7,11,13,17
We get this result: 
Balanced: 1,13
More weigth on side A: 17,19,23
More weight on side B: 5,7,11

Then we get here only two or three balls in each group. 

Third measure:

One of this balls we get in the second measure should be the inmediate superior in order to the unsorted ball. To determine which of them:
  • If the second measure was balanced the group of possible places of the unsorted balls is reduced to (1,13). Here we can include also the possible position 25 in the scale of order. To find which one of the three in this group or in the other two, the procedure is the same. 
  • We will just take the first ball of the three in ascendent order and we add the ball labelled with the difference between the first and the second ball of the group of three. On the other side of the balance we place the unsorted ball.
For ex. if the second measure has pointed out the group (17,19,23), we place  17+2 vs unsorted.
17+2 weights more, means unsorted is 17
17+2 is balanced with unsorted means unsorted is 19
17+2 weights less, means unsorted is 23.


To solve the other groups of 8 balls of the first measure the way is the same. 
For the group (6,8,9,10,18,20,21,22) we use as second measure: 
A: 6,8,21,22  B: 9,10,18,20

For the group (2,3,4,12,14,15,16,24) we solve with 
A: 2,3,16,24  B:4,12,14,15

Applying the third measure as described before to the resultant groups of two or three balls, it is possibile to determine te place of the unsorted ball in the scale.  

Edited on July 30, 2016, 10:47 am
  Posted by armando on 2016-07-30 04:34:38

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