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 Comparing weights (Posted on 2014-09-09)
A certain Mint manufactured 4010 silver dollars.
It is known that if any two coins were weighed against each other the difference in weight would be less than 1/100 of an ounce.

Prove that the coins can be partitioned in such a way that the total group weights differ by less than 1/100 of an ounce.

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 The Noah's Ark method (spoiler) | Comment 1 of 3
Add coins two at a time, one on each side of a balance scale.  Continue until there are 2005 coins on each side of the scale.

Before each addition, compare the next two coins to be added.  Add the heavier coin to the lighter pile.  Of course, if the two piles are equal at any point, you could can add the next two coins without comparing.  Also, of course, if the next two coins are of equal weight, either one can be added to the heavier pile.

By induction, it is obvious that the difference in weight between the two piles is less than 1/100 of an ounce before and after each pair of coins is added.  If adding two more coins does not change which pile is heavier, then it has reduced the amount of the difference (or at worst, has left it unchanged).  And if the addition of two more coins makes the heavier pile lighter, then the difference is necessarily less than 1/100 of an ounce.

 Posted by Steve Herman on 2014-09-10 09:08:49

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