Eight pills are on a table.all of the same color,shape,taste,and size.
However, one is poisonous and weighs a fraction more than the others.
Your job is to identify the poisonous pill.
You have a balance scale you can use, but you can oly use it twice.
How can you be sure to find the poisonous pill in these two tries?
This type of problem is really an extension of the typical problem of determining, through the use of a simple balance scale, which of several items is the odd one of the lot, knowing what characteristic makes it odd in the first place. For instance, a typical "find the odd item" problem is
You have eight balls, all identical in appearance, but with one ball weighing slightly more than the other seven. You can't tell which is the heavier by holding them in your hand, or rolling them, or bouncing them. You have a simple balance scale at your disposal. How can you find out which ball is the heavier in two weighings on the scale?
Notice you are told that the odd item, the ball in this case, is "heavier" than all the others. The procedure to follow is as follows:
1--Place three balls on each side of the balance scale.
2--If the scale balances, the heavy ball is one of the two still on the table and weighing those two against each other will tell you which is the heavier.
3--If the scales don't balance, the down side of the scale contains the heavy ball.
4--Take two of the three balls on this down side from step 1 and weigh them against each other.
5--If they balance, the third ball of the 3 from the original step 1 down side is the heavy ball.
6--If they don't balance each other, the ball on the down side is the heavy ball.
This approach also works with 9 balls, differing only slightly from the 8 ball case.
1--Place three balls on each side of the balance scale.
2--If the scale balances, the heavy ball is one of the three still on the table.
3--Weighing two of those on the table against each other. If they balance, the third is the heavy one.
4--If they don't balance, the one on the down side is the heavy one.
5--If the scales don't balance in step (1), the down side of the scale contains the heavy ball.
6--Take two of the three balls on this down side from step (1) and weigh them against each other.
7--If they balance, the third ball of the 3 from the original step (1) down side is the heavy ball.
8--If they don't balance each other, the third ball is the heavy ball.
This problem can be generalized in order to handle any number of items. If you know whether the item is heavier or lighter at the outset, you can determine the odd item amongst 3^n items in n weighings (^ means raised to the power of). Thus, a problem involving 3 to the n power items can be solved in n weighings.
1--Place 3^(n-1) items on each side of the balance scale which will immediately tell you which group of 3^(n-1) items contains the odd item.
2--Knowing which group of 3^(n-1) items contains the odd item, divide that group into three equal groups of 3^(n-2) items each, and place one of each of these groups on each side of the balance scale which , again, immediately tells you which of the three new groups contains the odd item.
3--The continuation of this process of reducing the size of the groups by two thirds at each successive weighing will ultimately leave you with the knowledge of, the one and only, odd item.
Take, what might appear to be, an outlandish problem as an example. In how many weighings on a simple balance scale can you determine which of 243 coins is heavier than all the others? I could have selected 729, 2187, 6561, etc., but 243 will get the point across.
1--243 = 3 to the 5th power so 5 weighings will tell you the odd coin.
2--Place 3^4 = 81 coins on each side of the scale leaving 81 on the table.
3--If they balance, the third pile contains the heavy coin and if they don't balance, the side of the scale that went down contains the group with the heavy coin in it.
4--Taking the known group with the heavy coin in it, divide it up into three groups of 3^3 = 27 coins each and place one group on each side of the scale.
5--Following the same logic of step 3, remove the group with the heavy coin in it and put the others aside.
6--Having the group with the heavy coin in it, divide it up into three groups of 3^2 = 9 coins each and place one group on each side of the scale.
7--Following the logic of step 3 again, remove the group with the heavy coin in it and put the others aside.
8--Divide this group up into three groups of 3^1 = 3 coins each and place one group on each side of the scale.
9--Following the logic of step 3 again, remove the group with the heavy coin and place the others aside.
10--Divide this group up into three groups of 3^0 = 1 and place one on each side of the scale.
11--You now know for certain which coin is the heavy coin.
