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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.

  Submitted by Dennis    
Rating: 5.0000 (1 votes)
Solution: (Hide)
Let x(k) represent the weight of a coin in bag #k. --> x(1)+x(2)w+x(3)w^2+x(4)w^3+x(5)w^4=2800. Since the weight of every coin is at least one, 1+w+w^2+w^3+w^4 < 2800 --> w < 7. Since the weight of every coin is at most w, 2800 < w+w^2+w^3+w^4+w^5 --> w > 4. So w=5 or 6.

If w=5 then x(1)+5x(2)+5^2x(3)+5^3x(4)+5^4x(5)=2800 --> x(1)=0 mod 5 --> x(1)=5 --> x(2)+5x(3)+5^2x(4)+5^3x(5)=559 --> x(2)=4 mod 5 --> x(2)=4 --> x(3)+5x(4)+5^2x(5)=111 --> x(3)=1 --> x(4)+5x(5)=22 --> x(4)=2 and x(5)=4.

If w=6 then, using a similar analysis, x(1)=x(2)=4 so condition c) is violated.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionPuzzle SolutionK Sengupta2022-07-29 01:45:07
re: solutionArt M2007-02-22 03:30:32
re(2): Need better definitionhoodat2007-02-21 23:44:44
re: Need better definitionDennis2007-02-21 20:25:57
QuestionNeed better definitionhoodat2007-02-21 17:41:16
SolutionsolutionCharlie2007-02-21 17:01:05
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