A group of friends find a bag with 56 coins. They do not know very well how to distribute them since they cannot do it in equal parts so one of them comes up with a solution:
Each of us will throw a coin in the air, if it goes expensive, it will take 5 coins and if it goes cross, it will take 6 coins.
Once the distribution was over, there was no money left over and we know that among those who received 6 coins there were two brothers.
How many of them received 5 coins and how many of them received 6 coins?
If we call X to the number of friends who received 5 coins and Y to the number of friends who received 6 coins, we get the following equation: 56 = 5x + 6y
Since we know that X and Y must be integer values, we can try different values of X until we get an integer value of Y considering that as stated in the statement there are at least two people with 6 coins.
So the solution is that 4 friends received 5 coins, and 6 friends received 6 coins.