Four married couples, with name A, B, C, D (male) and E, F, G, H (female), meet for a game of chess. They form four groups of two players. The following information is given :
- B plays against E
- A plays against C's wife
- F plays against G's husband
- D plays against A's wife
- G plays against E's husband
The wife of B is ...

Note by Jansen Wu
2 years, 7 months ago

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Here is another solution:

Since A plays against C's wife, and D plays against A's wife, then B could only plays against B's or D's wife.

If B plays against his wife, which mean E is B's wife, then G plays against B, which contradict.

That means B plays against D's wife, therefore D is E's husband. And because G plays against D (E's husband) that mean G is A's wife, F plays with A so F is C's wife, then the remaining B's wife is H

- 2 years, 4 months ago

Do case by case: Start by showing that A cannot be F's husband.

- 2 years, 7 months ago

How can I show that ?

- 2 years, 7 months ago

Actually you can start from anywhere.

Basically what I'm doing is case by case:

C's wife is either E, F, G, or H.

Case 1: Suppose C's wife is E, then from statement 2, A plays against E. But from statement 1, B also plays against E. Which is a contradiction.

Case 2: Suppose C's wife is F, then from statement 2, A plays against F, And from statement 3, A is G's husband. And from statement 4, D plays against G, This leaves us with C plays against H. And everything checks out.

Case 3: Suppose C's wife is G, then from statement 2, A plays against G. From statement 3, F plays against C. This leaves us with D plays against H. So A's wife is H. But because G plays against A, then A is E's husband. Which is a contradiction.

Can you work out Case 4 yourself?

- 2 years, 7 months ago

I see, thank you very much

- 2 years, 7 months ago