A wise sage gives you eight pennies and a two pan balance. One penny is heavier than the other seven. What is the least number of weighings that are required to determine which penny is the heavy one?
Solution
Remove two pennies from the pile and put them aside.
Put three pennies on each pan.
If one side is heavier than the other, put the light pennies aside. Put one of the remaining pennies aside.
Put one of the two remaining pennies on each pan. If one side is heavier that the other then you have found the heavy penny. If they are both the same then the penny you just put aside in step 3 is the heavier.
If both sides weigh the same in step 2 then one of the pennies you put aside in step one is the heavier.
Place one of those pennies on each pan. The heavier side is the penny you are looking for.
If you counted the number of weighings for either case 3 or case 5 you will find that there are only two weighings required.
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