Answers to last issue's Brain Teasers

1. Coloured weights: This is a variation on an old problem. You have six weights. One pair is red, one pair is blue and one pair is white. In each colour one weight is a little heavier than the other but they look exactly alike. All the heavier weights (in each colour) weigh the same and so also the lighter weights.
Using only two separate weighings on a balance, how can you identify which is the heavier weight of each pair?
Ans: Now, if we could weigh three times, we can simply weigh one weight of each colour against its partner and then we'll know which is the heavier weight in each colour. That is, if we put the two blue weights on either pan of a balance, we will know which is heavier. But we are allowed only two weighings. So we must clearly weigh more than one weight at a time. So let us put a blue in each pan and also a red weight in the left pan and a white weight in the other. There are three possibilities: they weigh the same; they weigh different: the left pan is heavier, or the right pan is heavier.
(a) Suppose the pans balance: Now we know that the two blue balls have different weights (we only don't know which is heavier). In fact, the weights of each colour weigh different amounts. So if the two pans balance, that can only mean that the lighter blue is paired with the heavier one of the other colour (red or white) and vice versa. We still don't know which blue is heavy or light, but the ONLY way the pans can balance is if the red and white weights are different.
So all you have to do is to weigh these two against each other to find out which is the heavier one. If it turns out that the red one is heavier, then its partner of the same colour is lighter. Also it means the white one is lighter so that its partner is heavier. Also, we remember which blue weight the red one was weighed with, and so if this red were heavier, that blue weight (kept in the left pan in the first weighing) is lighter. Of course if the red one turns out to be lighter, all the answers are the other way, but in either case with two weighings you can identify the heavier weights.
(b) Suppose the pans did not balance: Suppose the red-blue combination was heavier than the white-blue combination in the first weighing. Under what conditions can this happen? The key point to realise is that weights of the same colour can never be the same: one is light and one is heavy. If you think of the possibilities, you will see that this can only happen if the blue weight was heavier in the left pan. See the table where the possibilities for the red and blue weights in the left pan are shown against the possibilities for the white and blue weights in the right pan (and also case (a) in the last line).
Now we know that the left blue weight is heavier. But what about the others? Do we know anything about the red or white colours that we have weighed along with the blue ones?
Looking at the table, we see that there are two possibilities: (i) either the red and white weights are equal or (ii) the red is heavier. This also means that of the red and white weights that were NOT weighed, either they are of equal weight or the red is lighter!
Now take the red from the red-blue combination and the white from the white-blue combination. Weigh these together against the remaining red and the remaining white. If possibility (ii) were true, the red is heavy and the white is light, and you will weigh it against the remaining two weights which is a light red and a heavy white. So the pans will balance. So in the second weighing, if we see that the pans balance, we know that the red which was weighed in the first round is heavy and the remaining white is heavy, so we have found all the heavy weights.
Suppose possibility (i) is true. Then the weights (red and white) are either both light or both heavy. In the second weighing (against the remaining red and white weights) the pan will not balance. The heavier side will be the one containing both the heavy white and the heavy red.  So in this case also we can indentify all the heavy weights with just two weighings.

2. Arrange 10 Re. 1 coins (or any identical coins) in a triangle or bowling pin format as shown. In your mind, join the centres of each coin to each other. You can see several shapes, including several equilateral triangles. What is the smallest number of coins you must remove so that you will not be able to draw an equilateral triangle of any size with the coins as centres?
Ans: Let the coins be labelled A, B, C, ..., H, I, J as shown in the figure. Let's first try removing just one coin. If we remove only coin A, there are still many equilateral triangles left, such as BGI, etc. The same is true if you remove coin E.  You can try with any of the coins and you'll find that there are still equilateral triangles left. 
Now try removing two coins at a time: say A and E. But DGH is still left, etc. By now you can see that unless you remove ALL the coins in the central triangle, that is, E, H, and I, you will always have an equilateral triangle left. Of course you could have removed (D,B,E) or (C,E,F) instead. So we have to remove at least these three.  What do we have left? The outer equilateral triangle AGJ! In order to remove this, we must remove the edge coin, A. So removing A, E, H, and I leaves no more equilateral triangles as shown. So the least number of coins you need to remove is four.
This puzzle was given by the famous mathematician, writer and puzzle-creator Martin Gardner and the solution was first found by a Japanese person called Kobon Fujimura.