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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?**

#### Solution

If we call * X* to the number of friends who received 5 coins and

*to the number of friends who received 6 coins, we get the following equation: 56 = 5x + 6y*

**Y**Since we know that * X* and

*must be integer values, we can try different values of*

**Y***until we get an integer value of*

**X***considering that as stated in the statement there are at least two people with 6 coins.*

**Y**So the solution is that **4 friends received 5 coins, and 6 friends received 6 coins**.