Help me please

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 ...

#Logic

Note by Jansen Wu
5 years, 6 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

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

Tran Hieu - 5 years, 3 months ago

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

Pi Han Goh - 5 years, 6 months ago

Log in to reply

How can I show that ?

Jansen Wu - 5 years, 6 months ago

Log in to reply

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?

Pi Han Goh - 5 years, 6 months ago

Log in to reply

@Pi Han Goh I see, thank you very much

Jansen Wu - 5 years, 6 months ago
×

Problem Loading...

Note Loading...

Set Loading...