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?

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**