#### I found this puzzle in the internet. It looks quite easy to solve, but if you want to dig deeper into this you can find quite interesting fact.

**Solution (without spoilers):**

Let's name those rules with letters as follows:

**A**206 twoNumbersAreCorrectButWrongPlaced

**B**682 oneNumberIsCorrectAndWellPlaced

**C**614 oneNumberIsCorrectButWrongPlaced

**D**780 oneNumberIsCorrectButWrongPlaced

**E**738 nothingIsCorrect

How many codes is still in game after applying rules?

Count valid codes in case we are applying only two rules:

X | A | B | C | D | E |

A | 69 | 14 | 25 | 25 | 42 |

B | 14 | 192 | 40 | 40 | 72 |

C | 25 | 40 | 386 | 144 | 152 |

D | 25 | 40 | 144 | 386 | 78 |

E | 42 | 72 | 152 | 78 | 343 |

Now let's focus on AB and another rule:

**ABA**14

**ABB**14

**ABC**1

**ABD**7

**ABE**10

So if the winner exists we can focus on rules ABC to find it.

If you want to figure out how to solve it in java you can find answer on github [WARNING: spoiler in tests]: