You have to sort five weights weighing 51g, 52g, 53g, 54g, and 55g. You have a balance scale with which you can compare the weights. But after solving so many sorting puzzles, it is starting to break down.
If the difference between two weights is greater than 1.5g, the scale will correctly determine which side is heavier. If the difference between the weights is less than 1.5g or equal, the scale will indicate the weights are equal.
Sort the weights in the smallest number of weighings.
Being unable to follow Charlie's explanation (maybe I was too tired)...
My inferior method requires at most 7 weighings, 1 more than Charlie's. (Assuming Charlie is right, I did not mark this post as a Full Solution).
Any four of the five weights, weighing two against the other two, will result in two inequalities and one equality.
e.g.
51+53=104<54+55=109
51+54=105<53+55=108
51+55=106=53+54=107 (106=107 on the broken scale)
So three weighing will accurately sort any four of the five weights.
If the weights are A,B,C,D,E, then let 3 weighings sort four of them:
A<B<C<D
3 more weighings will sort B,C,D,E
If E<B<C<D
then 1 more weighing: E against B
if E=B, then the order is A<E<B<C<D
if E<B, then E<A<B<C<D
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Posted by Penny
on 2004-08-03 06:14:07 |
Inferior alternative
