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 A few coins (Posted on 2004-11-19)
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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 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 = 150Or, x/8 <= 150/8 = 18.75Or, 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+26 = 1+511 = 1+107 = 2+512 = 2+ 1015 = 5 + 10We 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........No6.......48........No11......88........No7.......56........No12......96........No15......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+50and, 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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