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Weighing Coins (Posted on 2007-02-21) Difficulty: 3 of 5
There are five bags of coins such that:

a) the weight of any coin (in ounces) is a positive integer less than or equal to w
b) all coins in a given bag weigh the same
c) the weight of a coin in bag #1 is different from the weight of a coin in bag #2
d) when one coin from bag #1, w coins from bag #2, w2 coins from bag #3, w3 coins from bag #4, and w4 coins from bag #5 are combined and weighed, the result is 2800 ounces

Determine analytically, the weight of a coin in each of the five bags.

See The Solution Submitted by Dennis    
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Solution solution | Comment 1 of 5

The weight of each coin can be considered a digit in a variant of base-w representation.  In the usual base-w representation, digits can be from 0 to w-1, but in this variation, they go from 1 to w.

The usual base conversion method is to divide the number by the base. The remainder becomes the first (or next) digit, placed to the left of any preciding digits found, and then using the quotient as the new dividend for the process to continue.

In this case, we must convert any zero remainders to w. At the same time, we must subtract 1 from the quotient before using it as the next dividend, to account for the "borrowing" done.

As the powers of the base go from zero to four, we need a 5-digit number in the peculiar base-w system to represent 2800 decimal.  So we can try various bases.  The following table shows this base-w representation for various w:

 2           12122221112
 3           3211131
 4           223234
 5           42145
 6           16544
 7           7777
 8           5358
 9           3751
10           279a
11           2116
12           1754
13           1375
14           e3e
15           c6a
16           aeg
17           9bc
18           8ba
19           7e7
20           6jk

Note that a=10, b=11, c=12, etc. in decimal representation.

So for the base w being 5 or 6, we get a 5-digit representation of 2800 decimal.

As the lower powers of w are on the right in interpreting the numbers, bag 1 is represented by the rightmost digit and bag 5 by the leftmost.

When the base is 6, the rightmost two digits are the same, and thus bags 1 and 2 would have the same weight coins so this possibility is disqualified.

So the weights of the coins in bags 1 through 5 are 5, 4, 1, 2 and 4 respectively.

The following program tries the bases, b, from 2 to 20, resulting in the above table:

n = 2800
FOR b = 2 TO 20
 m = n
 ns$ = ""

   d = m MOD b
   m = m \ b
   IF d = 0 THEN d = b: m = m - 1
   ns$ = MID$("0123456789abcdefghijklmnopqrstuvwxyz", d + 1, 1) + ns$
 PRINT b, ns$


  Posted by Charlie on 2007-02-21 17:01:05
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