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Xerxes has 6, 7 and 8-arm robots in his spaceship. Unfortunately, the 7-arm robots are misprogrammed and everything they say is a lie. The other robots are fine, they always speak the truth.

Once Xerxes hears a conversation from 4 robots, but without seeing them. They talk about their arms:

The first robot claims: "We 4 have 28 arms together."

"No", says the second, "it's 27."

"Not true, 26," says the third.

"Wrong, it's 25", says the fourth and ends the conversation.

How many arms does the 4th robot have?

(A) 6
(B) 7
(C) 8
(D) 6 or 8

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    $\begingroup$ Is the second robot making a statement by saying "No"? A statement: "No, it's not true that we have 28 arms together" which can be either true or false? The same for other robots saying "Not true" and "Wrong"? $\endgroup$
    – Przemysław Czechowski
    Jun 28 '19 at 15:21
  • $\begingroup$ @PrzemysławCzechowski: it's the same thing, the second robot is saying "No [we don't have 28 arms], we have 27". You're overthinking the verbal aspect, it's equivalent to R1 saying "Our total arms is 28", R2 says "27", and so on... regardless whether they say they explicitly disagree with each other or not, their stories implicitly disagree. $\endgroup$
    – smci
    Jun 30 '19 at 18:54
  • $\begingroup$ Hey, wait a minute. The first robot says "We 4", right? But that is the truth! $\endgroup$
    – Mr Lister
    Jun 30 '19 at 19:21
  • $\begingroup$ Oh you're right. But still the full statement is wrong. $\endgroup$
    – Matti
    Jun 30 '19 at 20:14
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    $\begingroup$ Does this answer your question? King Octopus and Servants Puzzle $\endgroup$
    – Hemant Agarwal
    Apr 26 at 12:18
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Answer

(B) 7

Reasoning

They all disagree so there can be, at most, one robot telling the truth. This means that at least 3 of the robots have 7 arms, making the overall total number of arms either 27, 28 or 29. Hence, the last robot is lying.

Furthermore

The overall number of arms cannot be 28, otherwise all robots have 7 arms and the first one tells the truth.
The overall number of arms cannot be 29, otherwise the robot with 8 arms has told a lie.
Hence, the overall number of arms is 27 and the second robot has 6 arms.

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    $\begingroup$ Was just typing a solution up. Well done with the quick answer! $\endgroup$
    – gabbo1092
    Jun 27 '19 at 14:13
  • $\begingroup$ @gabbo1092 Thank you. $\endgroup$
    – hexomino
    Jun 27 '19 at 14:15
  • $\begingroup$ I hope this logic riddle wasn't too easy.(?) $\endgroup$
    – Matti
    Jun 27 '19 at 14:23
  • $\begingroup$ @Matti Not when SHODAN can help you solve it. $\endgroup$
    – rinspy
    Jun 28 '19 at 16:09
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I think, all the other answers (though correct) are way to complicated: of course the right solution is

He has 7 arms

But the reasoning is much easier:

The number 25 of the last robot (and thats the one we are interested in) can only be true, if there are 3* 6 arms + 1*7 arms. But then 3 robots tell the truth.... as they all give different answers this is not possible: the last one lies...

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B - seven

After round one, we know that the robots have one of:

$6688, 6868, 6886, 6778, 6787, 6877$
$7666, 7667, 7668, 7677, 7686, 7688, 7766, 7767, 7768, 7776, 7778, 7866, 7868, 7877, 7878, 7886, 7887, 7888$
$8668, 8686, 8866, 8776, 8767, 8677$
arms.

After round two, we know:

$6778, 6787, 7668, 7677, 7686, 7766, 7768, 7778, 7866, 8776, 8767$ arms.

After round three, we know:

$6778, 7677, 7766, 7778, 8776$ arms.

After round four, we know:

$7677$ arms.

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  • $\begingroup$ why isn't 6778 (and several other setups where the first digit is 6 or 8, and they sum up to 28, but there are some 7s) possible after the first round? $\endgroup$
    – elias
    Jun 30 '19 at 17:13
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    $\begingroup$ @elais, thanks, tixed it! $\endgroup$
    – JMP
    Jun 30 '19 at 18:04
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Since all four robots give a different answer, they must either be all lying, or one is telling the truth and the rest are lying.

From the first statement:

The first robot claims: "We 4 have 28 arms together."

we can conclude that:

All four robots cannot be all 7-armed robots, as that would be a true statement if they were all 7-armed, and 7-armed robots can't tell the truth. So now we know that one is telling the truth, and the other three are lying 7-armed robots.

And the logic follows that:

There is either one 6-armed robot mixed in with three 7-armed robots (for a total of 27 arms), or one 8-armed robot mixed in with three 7-armed robots (for a total of 29 arms).

From this, we can deduce that:

The non-7-armed robot has 6 arms, because one answer contains "27", but no answer has "29".

which reveals to us that:

The truth-telling robot (the one with 6 arms) is the second robot, because that answer matches the correct number of arms. All the rest (the first, third, and fourth) have 7 arms.

Therefore, the answer to the puzzle is:

(B) The fourth robot has 7 arms.

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B: 7 arms

because

there are 4 different statements, at least three are lies, all 4 cannot be lies because 4 x 7 = 28 would not be a lie. 3 x 7 = 21. the fourth robot must lie since there must be at least 27 total. the 6 armed robot answered 27, the other statements are lies.

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Answer is 7

The first robot claims: "We 4 have 28 arms together."

The above statement implies that all the 4 robots have 7 arms each and it is clearly mentioned that 7 arms-robot say lies, so obviously this statement is rejected on a single look.

Now the key point is-since only one statement can be true,hence all other three statements should be made by 7 arms-robot.

So what we need is to solve this easy equation-

7+7+7+x

we need to find our x such that it fits any one of the above statement . And in this case it is clear that our x should be 6 ,which marks our second statement as true. Hence the statement of the 4th robot is wrong and hence..

4th robot has 7 arms.

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    $\begingroup$ The first robot claims: "We 4 have 28 arms together." But that could be 6 + 6 + 8 + 8, or 6 + 7 + 7 + 8. So you can't immediately say that's a lie. $\endgroup$
    – Weather Vane
    Jun 27 '19 at 17:27
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To put it short,

- 3 robots are lying and are 7-armed: if more than one would tell truth, some answers would be same.
- If there were 4 7-armed robots, answer 28 should not come because it is true.
- So only one other 6 armed robot can be and there is 27 together, so he is the second.
- The last one has 7.
There is a small mistake in the task. 4th says "Wrong, it's 25" but negation "wrong" is true.

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3 short answers....

Either all 4 are liars or 3 are lying. So the combinations are 7,7,7,6(27) or 7,7,7,7(28) or 7,7,7,8(29). The 4th robot said 25 so he must be a liar and so has 7 arms. Easy. You dont need to work out anything else.

Or

The only way to get 25 as an answer so robot 4 is telling the truth is 6,6,6,7 which would mean 2 others are telling the truth too, and they cant be if they give different answers. So hes a liar and must have 7 arms. Even easier?

Or

If you ask yourself if you can tell if 4 is telling the truth at the start and ignore everything else you can basically find that he has to be a liar for any others to be liars which they have to be.

So

Ive gone and taken all the fun out of it. Sorry. Lol

But

But as Przemysław Czechowski says, really the logic is flawed because it says that the 7 arm robots say only lies. Well, they both tell the truth and lie by saying "not true". So the whole puzzle disappears into its own butt in a flash of logical illogicality. Infact number 4 is telling the truth by saying not true which means he is broken along with my spirit. But he also isnt because its also really not true in this case. Oh dear, what a terrible mess. I thimk its safer to travel back in time and become your own father than this(and that is obviously impossible as you become your own mother in the process and so the whole trip never happens to start with, but i digress)

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