The situation is this: on the table there are 10 coins, 5 shot on the side of the head And 5 shot on the side of the cross. Let’s imagine, however, that the light is off or you are blindfolded, and therefore it cannot see which coins are shot on the side of the head and which on the side of the cross. The coins are totally indistinguishable, So even by touching them it is not possible to understand which are those with the head and which ones with the cross.
The request is this:
Can you divide the coins into two groups of 5, so that the two groups have exactly the same number of coins shot on the head and the same number of coins shot on the side of the cross?
You are given the possibility of touch the coins and of turn them How many times do you want to remove the bandage before and find out if you managed to solve the riddle. What are the moves you need to do?
The solution to the tidy of the 10 coins
At first glance, this riddle seems absolutely impossible! At the level of chancedividing the 10 coins into two groups of five, could we have taken any combination of heads and crosses, and therefore how to divide them in order to make the two groups identical?
The first observation that must be made is that, being both the number of heads that the number of crosses a odd numberin any way we will divide them into two groups cannot be in the same number. Precisely because, in fact, it is an odd number. But we know that we can coins How many times we want, so as to change the number of heads and crosses. But we are blindfolded! So how to do it?
The solution comes from a reasoning that does not immediately jump to the eye (or to the brain), but which is very simple! The answer lies in the fact that dividing any 10 coins into two groups, We can be sure that the number of heads in a group will be equal to the number of crosses in the other and vice versa.
Let’s try to understand it better. Let’s imagine we have divided the 10 coins into two groups of any 5 and call x the number of heads in the first group. Since there are 5 total heads, we can be sure that in the second group the number of heads will be 5-X, given that (5-X)+(x) = 5which is the total number of heads. However, each group has just 5 coins, so if in the first group there are x heads, they will surely be there 5-X crosses! And equally, if in the second group there are 5-X heads, they will surely be there x crosses!
This means that, in any way the 10 coins are divided, the number of heads in the first group will be equal to the number of crosses in the second group and, equally, the number of crosses in the first group will be the same as the number of heads in the second.
Understood this concept, we found the solution: once divided the coins into two groupsit will be enough for us turn all the coins of one of the two groups To get to have the same number of heads and crosses in the two groups, thus obtaining two identical groups!
