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.

Per the wording of the problem, there are multiple solutions.  By the wording, bags 3-5 could contain coins weighing the same as either bag 1 or bag 2.

Assuming that all bags weigh differently,  there are still many combinations that work.  For example:

Bag 1:  6
Bag 2:  4
Bag 3:  5
Bag 4:  2
Bag 5:  1

w = 6.65165188040866 . . .

Is there something I am missing here?

Posted by hoodat on 2007-02-21 17:41:16