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A few coins (Posted on 2004-11-19) Difficulty: 1 of 5
In Levikland, there are coins worth 1, 2, 5, 10, 20, 50 and 100 perplexii. A has twice as much money as B, who has twice as much as C, who has twice as much as D. How can this be, if everybody has two coins?

See The Solution Submitted by e.g.    
Rating: 2.9231 (13 votes)

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Solution Puzzle Solution Comment 14 of 14 |
Let the respective amount of money possessed by A, 
B, C and D be u, v, w and x (in perplexus currency)
Now, the amount possessed by C is half of that of D,
the amount possessed by B is half of that of C and
the amount possessed by A is half of that of B.

Thus, w = x/2, v = x/4 and u = x/8
But, x<= 100+50 = 150
Or, x/8 <= 150/8 = 18.75
Or, x/8< = 18, since x/8 is a positive integer.........(i)
Since each of the individuals must possess precisely 
two coins of distinct available denominations, it
follows that the possible values of x/8 satisfying
(i) are:
3 = 1+2
6 = 1+5
11 = 1+10
7 = 2+5
12 = 2+ 10
15 = 5 + 10

We will now check whether x is expressible as the sum of two distinct
available values for these values of x/8.
x/8.....x........Is x expressible as the sum of two distinct available 
                 values
3.......24........No
6.......48........No
11......88........No
7.......56........No
12......96........No
15......120.......Yes, 120 = 20 + 100
In terms of the above table, the only possible pair 
is: (u, x) = (15, 120)
Thus, w = x/2 = 60 = 10+50
and, v = x/4 = 30 = 10+ 20
Accordingly, each of w and v are expressible as the 
sum of two distinct values. This is in conformity
with the provisions of the problem.
Consequently, the distribution of the coins is 
possible if the respective amounts possessed by
A, B, C and D are 15 perplexii, 30 perplexii,
60 perplexii and 120 perplexii.

Edited on October 24, 2007, 1:35 pm
  Posted by K Sengupta on 2007-10-20 00:24:11

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