Conference rusure::math

121.0. "You can't trust anyone" by NY1MM::PINSLEY () Wed Aug 08 1984 10:20

A relatively simple logic problem (as compared with some of the entries I've
read in this notes file).

A king has in his kingdom 10 men who mint gold coins.  Every mint man makes
the same type of gold coin (i.e. the same size and weight).  One day the
king discovers (through his newly established spy network) that one of
his trusted mint men is pilfering gold from his coins.

In fact, his information reveals that this one dishonest man is producing
hollow coins that look identical to the real thing - but are exactly 1/2
the officially accepted weight.

In order to get to the bottom of the problem, the king orders all 10 mint
men to the castle and tells each to bring a bag of 500 sample coins.  At
the kings disposal is a spring scale (i.e. one that gives a numeric reading -
not a balance scale) and a wise man.

The king knew that he could weigh one coin at a time from each man and
consequently determine the identity of the thief.  He chose instead to test
the wisdom of his so-called "wise man" (who had recently fallen from grace
with the king for reasons not publicly known).  The king ordered the wise
man to determine the identity of the thief beyond any doubt by using the
scale only once (i.e. getting one numeric reading).  He was permitted to
take as many coins from any or all of the mint man to accomplish the task.

Upon pondering his dilemma, the wise man gathered up some coins, placed them
on the scale and identified the thief.  How did he do it?

By the way, the wise man had no choice but to succeed.  He was also wise enough
to realize that failure would have resulted in the loss of his wise head.

If one considers the weight of a correct coin one scale unit,
then:

If each man is called 1 thru 10, and each man places twice
his called number of coins on the scale (i.e. man 1 places 2 coins,
man 2 places 4, thru 20 coins from man number 10) the difference 
in coin units from the supposed correct value is the man number.

If I am not incorrect, this is a variation of a drugist mixed pill
problem using a balance beam I heard as a little boy.

The previous reply is correct.

I told you it was easy.  The key is to realize that in order to distinguish
the mint men, a different number of coins must be taken from each.