Truth-Seeking a knight, knave or joker by Brilliant Staff

I didn't solve this problem the way the Solution shows. I just read one statement 'They all know what each other is (a knight, knave or joker)' If they all know what each other is then if there were multiples of any of them there would only be two options showing! There must be one knight.

#Logic

Easy Math Editor

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Comments

You could have submitted your solution as a solution to the problem, right?

Agnishom Chattopadhyay - 3 years, 9 months ago

You can do that!

Stephen Powell - 3 years, 9 months ago

Unless it's by Brilliant, you can't submit a solution, otherwise, you can.

William Huang - 3 years, 8 months ago

@William Huang – Thank you!

Stephen Powell - 3 years, 8 months ago

Actually, this is technically wrong. Because of the word "or," there might still be, suppose two knights and a knave, proving your solution false. However, that is quite a loophole and still a valid report because of possible amiguation.

Ayush Kumar - 3 years, 8 months ago

I solved this by assuming that Ellis was the one and only Joker. If Ellis told the truth, Farin would also be a joker so this isn’t a valid choice. If Ellis lied, Farin could be a knave or a knight. If Farin is a knight, that would make Gobi a joker so this isn’t a valid choice if Ellis is the one and only joker. If Farin is a knave, that would make Gobi a knave or a knight. If Gobi is a knave, this would make Ellis a knave or a knight which is not a valid choice. This leaves the option of Gobi being a knight which confirms Ellis is the joker. I believe if there is only one joker, there is one knave and one knight.

d k - 2 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$