### Random Post

### Recent Posts

### Recent Comments

Jim Randell on Enigma 1691: Factory part… | |

Jim Randell on Puzzle 52: Football on the Isl… | |

geoffrounce on Enigma 1691: Factory part… | |

Hugh Casement on Enigma 1070: Time to work | |

Jim Randell on Enigma 1070: Time to work |

### Archives

### Categories

- article (11)
- enigma (1,157)
- misc (2)
- project euler (2)
- puzzle (40)
- site news (44)
- tantalizer (42)
- teaser (3)

### Site Stats

- 177,808 hits

Advertisements

This is a similar problem to

Enigma 36andEnigma 183(both also set by Martin Hollis).Here is the setters solution:

And here is my programmed solution. This Python program uses the

filter_unique()function from theenigma.pylibrary. It runs in 44ms.Solution:Mrs. Um has 4 children. Mrs. Er has 5 children.And here is my analytical solution. (Which is basically what the program is doing).

To start with each mother knows how many children they have themselves (I hope), and we are told that in total they have 9 or 10 children.

So if U has 9 children, she knows that E must have 1 child. We denote this: U: 9 → 1.

And if U has 1 child, she knows that must have either 8 or 9 children. We denote this: U: 1 → 8,9.

The full list of possibilities is:

[0] Initially:

[1] U asks a question, so U does not know for sure how many children E has, so U cannot have 9 children, we remove the possibilities where U has 9 children:

[2] E asks a question, so E does not know for sure how many children U has, so E cannot have 1 or 9 children:

[3] U asks a question, so U does not know for sure how many children E has, so U cannot have 1 or 8 children:

[4] E asks a question, so E does not know for sure how many children U has, so E cannot have 2 or 8 children:

[5] U asks a question, so U does not know for sure how many children E has, so U cannot have 2 or 7 children.

[6] E asks a question, so E does not know for sure how many children U has, so E cannot have 3 or 7 children.

[7] U asks a question, so U does not know for sure how many children E has, so U cannot have 3 or 6 children.

[8] E asks a question, so E does not know for sure how many children U has, so E cannot have 4 or 6 children.

[9] U does not ask a further question as she now knows that E has 5 children.

But I don’t think E can know, at this point, how many children U has (they must have either 4 or 5).

However, we are now told that at least one lady has an even number of children, so U must have 4 children.

Interested that this puzzle is counter-intuitive. I’m glad that I reached the answer 4,5. Statement [9] seems right, and U only has 4 children since we are told that one lady has an even number of children. Could Professor Plato end with “with a DIFFERENT NUMBER OF CHILDREN, but 9 or 10 children between you.”

@Tessa: If Professor Plato had introduced them as having a different number of children then the pair (U=5, E=5) is not possible and so we would get:

[0] Initially:

[1] U asks a question. We remove the possibilities where U has only one choice (U=5, U=9):

[2] E asks a question. We remove the possibilities where E has only one choice (E=1, E=4, E=5, E=9):

[3] U asks a question. We remove the possibilities where U has only one choice (U=1, U=4, U=6, U=8):

[4] E asks a question. We remove the possibilities where E has only one choice (E=2, E=3, E=6, E=8):

At this point U would know that E must have 7 children, so wouldn’t ask a further question. E would know that U must have 2 or 3 children but without additional information wouldn’t know which case applies.

So we would only have a sequence of 4 questions.

Again if we are told that one of the ladies has an even number of children we can deduce at this point that U=2 and E=7.