Good scales have equal arms (arms are the the things that connect the actual scale to the center), but in one grocery stall, the arms of the scale are not equal. Pending replacement, the manager wonders if he can give correct weight this way:

"I'll balance a 1-pound weight on the left with sugar on the right, and then I'll balance the 1-pound weight on the right with some more sugar on the left, and the sugar will add up to exactly 2 pounds."

Will it? What are other (assuming that the above works, it may not) ways of weighing 2 pounds of sugar, if you also have a lead shot with you to help weigh? (Note and hint: The lead shot has an unknown weight. You can make it whatever weight you choose. Remember, the arms aren't equal, and you need 2 pounds of sugar.)

Lets call the ratio of the arms r.  Lets call the weight of the lead shot w.  The lead shot can be balanced with sugar to give sugar weighing w*r or w/r, depending on which side we place the lead.  The masses w*r and w/r can be used to get sugar weighing w*r^2, w, and w/r^2.  The balancing process can be iterated to generate all weights of sugar in the set S{. . ., w/r^3, w/r^2, w/r, w, w*r, w*r^2, w*r^3, . . .}.  Any weight can be constructed if it is a combination of the weights in the set S.

Posted by Brian Smith on 2004-05-24 10:09:51