Answers to last issue's Brain Teasers

1. There are 100 coins each in 10 bottles. Each of the coins weighs 10 gms. However, the manufacturers made a mistake and so one of the bottles contains coins that weigh only 9 gm each. How can you find out which bottle contains the defective coins by making just one weighing in a weighing machine?
Ans: Here it is not important whether one set is lighter or heavier than the rest: it is critical to know the actual weights of both the normal and the different sets of coins.
Of course if you weigh a coin from each bottle, you can find out which set is different but that needs 10 weighings and you have to do it in one go. Suppose you make the problem simpler: you have just two bottles full of coins and one set weighs 9 gms while the other weighs 10 gms. Now, if you take a coin from each set and weigh them together, you know the weight you will get is (10+9=) 19 gm. This will not tell you which is the lighter one, though! How can you improve upon this?
Suppose you took one coin from the first bottle and two from the second and weighed them together. What weight will you see? This will depend on which bottle has the lighter coins. There are two possibilities. If the first bottle had the lighter coins, then you have taken 1 lighter coin and two normal ones, so the weight you will see is (1X9 + 2X10) = 29 gm. If instead the second bottle had the lighter coins, you will measure a weight of (1X10+2X9) = 28 gm.
But now these two answers are different! So you can immediately tell which bottle had the lighter coins: if the weight is 29 gm, the lighter coins are in the first bottle and if the weight is 28 gm, they are in the second bottle. One way to understand this answer is to see that if both bottles had identical coins, you would have measured 30 gm. So, if the answer you get on actually weighing the coins is 1 gm less, then the first bottle has the lighter coins, and if the answer is 2 gm less they are in the second one.
Now we have solved the problem if there are just two bottles. How can we extend this result to 3, 4, ... 10 bottles? A little thought will tell you that the key to the problem is to weigh in one go, different numbers of coins from each bottle. Hence the answer is the following:
Take 1 coin from the first bottle, 2 from the second, 3 from the third and so on upto 10 from the 10th bottle. So you have (1+2+3+...10) = 55 coins in all. If all were the same weight, you would have got 550 gm. If the actual weight you get is 1 gm less than this, then the lighter coins are in the first bottle. If you get 2 gm less, they are in the second bottle and so on. You can see that it is important not really to know the different weights, but the amount by which the weights are different.
2. You have 12 balls identical in size and appearance but one is lighter or heavier than the rest.
You have a set of scales (balance) which cannot tell you the exact weight but can tell which is the heavier or lighter of two objects (the scale will tip over to the side of the heavier object or stay balanced if the two objects have the same weight).
Can you find out which is the odd ball with just three chances to weight them on the balance? Also, can you find out whether it's heavier or lighter than the rest at the same time?
Ans: This is yet another endless variation on the weighing problems. Each is slightly different and each one is nothing short of amazing in its inventiveness!
You do not know the weight of any of the balls, so clearly in this puzzle you have to weigh the balls against each other. There are 12, so let's start by weighing them 6 to a side. One lot is heavier and one lot is lighter. Now you don't even know if the odd ball is heavier or lighter, so we have come to a dead end fast! Clearly this cannot be the first weighing.
Let us instead try and weigh some of the balls. For instance, weigh 4 on a side, so take only 8 of them. To make it easier, let us call the set on the left side containing balls 1,2,3,4 as Set I and the ones on the right with balls 5,6,7,8 as Set II (so that the left-over ones, 9,10,11,12, are Set III).
There are two possibilities: either the odd ball is in Set I/II or it is in Set III. If it is in Set I/II, then when we do the first weighing, the scales will not balance and we will know that the odd ball is one amongst these eight. More importantly, we will know that any of the balls in Set III (9,10,11,12) are normal and can be used as a measure of the correct weight.
For simplicity, let us assume that Set I came out to be lighter than Set II. Since we do not know whether the odd ball is lighter or heavier than the rest, we cannot tell if it is in Set I or Set II. All we can say is that, if the odd ball is lighter, then it must be one of balls 1,2,3,4. Let us call them LN balls (either light or normal). If the odd ball is heavier, it must be one of the balls in Set II, i.e., 5,6,7,8 (let us call them HN balls, either heavy or normal). Set III balls are all normal, so we call them N.
To find out which is the odd one, then, we must weigh these balls against any one of the balls that we know are "normal" ones from Set III, let us say 9. Here is the trick that allows you to find the answer in three weighings: Suppose we weigh an LN ball against an N ball. If they weigh the same, the LN ball is actually normal, and if not, it is the odd (and light) ball. However, if we proceed this way, we may have to keep on weighing until we find the odd ball and we certainly cannot do this in 2 more weighings. We have to be smarter than this. So, in the second weighing, weigh not just one LN ball against an N ball, but in the following way: LN,LN,HN against LN,HN,N. For example, 1,2,5 against 3,6,9. We have therefore kept aside one LN ball (4) and two HN ones (7,8). Since we have a complicated combination, either side could be light or heavy, or they could balance. Let us examine each in turn.
(A) The two sides balance. So either 4 is the light odd ball or one of 7,8 is the heavy odd ball. Weigh 7 versus 8 (third weighing). If they balance the odd ball is 4 and it is lighter. If they don't balance, the heavier one is the heavy odd ball (since we know that they are HN, i.e., either heavy or normal).
(B) The two sides don't balance. Suppose 3,6,9 is heavier than 1,2,5. Now 3 can only be light/normal (LN) not heavy, while 5 can only be heavy/normal (HN) not light, so these can't be the odd balls. (Remember 9 is normal). Then ball 6 is heavier or either one of 1,2 is the odd light ball. Just as before, weigh 1 against 2 (the third weighing). If they balance, then 6 is the heavy odd ball; if they don't then the lighter of the two is the light odd ball.
(C) The two sides don't balance but now 3,6,9 is lighter than 1,2,5. Since 9 is a normal ball, this can happen if 3 is light or 5 is heavy To find which of 3 and 5 is the odd ball, weigh any one of them against a normal ball, say 9 (this will be the third weighing).
If Set I is heavier than Set II, you can repeat the entire argument above, interchanging the choices from Set I and Set II. 
Finally, we have the case that in the first weighing, Set I and Set II balance each other. Then the odd ball is in Set III and we can weigh any three of the balls in Set III against any three normal balls from the remaining sets. Repeat the arguments above carefully and find out that in this case also, we can not only find the odd ball in three weighings, but also determine whether the odd ball is heavier or lighter than the rest.