A distant planet is inhabited by creatures having more than two arms. The ones with an even number of arms are the knights who always tell the truth; the ones with an odd number of arms are the liars who always speak falsely.
The rules of the planet allow only creatures having eight, nine or ten arms to work as guards. Four guards are conversing amongst themselves:
- The magenta one says, "All together we have 39 arms."
- The cyan one says, "No, we do not."
- The brown one says, "The beige one has 8 arms."
- The beige one says, "The brown one is lying."
Determine the number of arms of each guard.
I agree with Charlie. I was put off by the multiple solutions and therefore did not post earlier.
a) Assume brown is a knight, then beige has 8 arms and is therefore also a knight. But beige says that brown is a liar, so this is a contradiction. Therefore our initial assumption is wrong and brown is lying (and has 9 arms). Beige must therefore be a knight of the 10-armed variety.
b) Magenta and cyan directly contradict each other, so one of them is a liar and one is a knight. This means that there are two liars and two knights, so the total number of arms must be even. So magenta is a liar (with 9 arms) and Cyan is a knight. There is no way to tell whether cyan has 8 or 10 arms. The total number of arms is either 36 or 38.
Magenta -- 9 arms
Cyan -- 8 or 10
Brown -- 9
Beige -- 10
I think this problem is difficulty 2, not 3.