This represents a circular balance.
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A light circular disk is needle point mounted. It has a graduated scaled track "o" from which identical measuring pans may be suspended.
Suggest the minimum number of pans as well as the minimum weighings needed, and the strategy required to determine which one of 9 coins is in weight variance with the other 8 in a visually identical collection. You should also be able to determine if the variant coin is lighter or heavier than each of the others.
At the end of each weighing, the coins being weighed will be on the balance, and the balance will be in equilibrium.
For additional discussion:
The set of nine has 7 regular coins with one weighing lighter, and the other weighing heavier than the others.
How might the circular balance assist?
Does a disparate difference in weight of the lighter to the normal than the heavier to normal influence the procedure?
[This thought was raised at review.]
(In reply to
Additonal commentary after reading the official solution (Spoiler) by Dej Mar)
It is indeed possible to accomplish this task with 3 pans
and 2 weighings as Daj Mar has indicated, one just has to ensure that the group
of 6 is weighed first.
<o:p></o:p>
Taking a step backward and readdressing a question posed by
Gamer: Minimum?<o:p></o:p>
Charlie did address this with equal spacings; and he suggested “with
the circular pan” when he certainly meant “circular disk.
Gamer also addressed this, taking to task minimum weighing as one
point, and minimum pans on the other hand.
At that point there was wording which I had not sufficiently addressed.
[I did note however that while he (Gamer) had the solution in hand, equilibrium –
original horizontal state of balance – had not been achieved. Unfortunately that was not made clear; my
assumption of equilibrium, yes faulty, that one’s action on a system should
return one to the original state.]
<o:p></o:p>
Definition of Minimum.
I think that from your “definition”
a count of six is the minimal value (3 pans in each of two weighings) is
required for surety before any action is taken? If we get lucky ten the count is 3?
I think I’ve addressed issues here; pity though that I
did not have the availability of what the first paragraph addresses to add
directly into the “official solution”.
Wonder if levik might however assist.
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Posted by brianjn
on 2008-03-29 08:28:31 |