Kalkulative Knaves and Knights
There are 6 people on the island, each who can either be a knight or a knave. A knight tells the truth, whereas a knave tells the lie.
Person 1: The number of either knights or knaves is the multiple of another of the different kind.
Person 2: There are less knaves than knights here.
Person 3: There is an odd number of knaves.
Person 4: The number of knaves is not a perfect square.
Person 5: The product of two numbers of knights and knaves is a perfect cube.
Person 6: The product of two numbers of knights and knaves is a perfect square.
What is the least possible number of knaves there?
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2 solutions
Why can't the answer be 1?
There would be 5 knights, 1 knave.
Everyone tells truth except P4.
P1 is correct; 5 is a multiple of 1.
P2 is correct; 1 < 5.
P3 is correct; 1 is odd.
P4 is lying; 1 is a perfect square.
P5 is telling the truth; 2 (number of a knight) • 4 (number of a knave) = 8, a perfect cube.
P6 is telling the truth; 1 (number of a knight) • 4 (number of a knave) = 4, a perfect square.
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Check P5 and P6 again.They're lying.
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How are they lying?
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@Siva Budaraju – If 5 knights, 1 knave,then 5*1=5 which is not a perfect cube or square.
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@X X – Oh, I interpreted the statement to mean that if you take the person # of one of the knights and the person # of one of the knaves, and multiply them, you get a perfect square/cube.
The wording is not very clear.
The problem can be worked out via number theory and careful observation. The reason why people are not all knaves nor knights is because of the first statement. The integer ends up dividing its complement since
6 − T ≡ 0 m o d T
is valid if T ≡ 1 , 2 , 3 , 6 . However, since T = 0 is also possible, then we count 5 possibilities in all, which also applies to F . It is very easy to see that the number 6 has the unique products with different properties. If T is the number of knights and F is the number of knaves, then
T + F = 6 → T + ( 6 − T ) = 6
But from Vieta's formula , T F must be an integer that automatically satisfies the quadratic equation. The discriminant should really help to observe the type of integer roots.
So the obvious basis is that T ≥ 1 , so that F ≤ 5 . But if the number of knaves is odd, then Person 5 must be lying by the nature of the number 6, forcing T ≥ 2 and F ≥ 2 . But this forces Person 6 to be telling the truth and Person 4 to be lying.
In this case, both numbers can't be odd. So we know that Person 1 and Person 5 are telling the truth. The rest should be logically simple.
Making general concepts most often eliminates more possible cases, simplifying them to the easier ones.
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I think what tripped me up is the wording of P5 and P6's statements. Could you change it to say, "The product of the number of knights among us and the number of knaves among us is a perfect square/cube"?
The way I read it, I thought it meant, "The product of the person number of someone who is a knight and the person number of someone who is a knave is a perfect square/cube."
@Siva Budaraju I think you misunderstood the meaning of two numbers of knights and knaves - what I feel it means is the total number of knights and the total number of knaves.
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Yes, I see that.
The wording of the problem should also be changed slightly, as right now the statements of p5 and p6 are ambiguous.
The answer could also be 4. Only persons 1 and 5 are knights while persons 2, 3, 4 and 6 are all knaves. We have two knights and four knaves so person's 1 statement is true as four is indeed a multiple of two. Furthermore, four times two is eight, a perfect cube which makes person 5's statement true as well. Eight, however, is not a perfect square hence person 6's statement is false.
Now with four knaves and only two knights, person 2 is lying as we in fact have more knaves than knights. It is also obvious that person 3 is lying as four is an even number, contrary to what he said about there being an odd number of knaves. Also four is a perfect square which makes person 4's statement false.
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P1,"The number of either knights or knaves is the multiple of another of the different kind."Though 4 is a multiple of 2,2 isn't a multiple of 4.
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It didn't say that it had to work both ways, just one way.
Otherwise, that statement would be the same as 'the number of knights and knaves are same.'
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@Siva Budaraju – Yes, the only possibility in which the statement works both ways is if there are as many knights are there are knaves (i.e. three of each), so @Siva Budaraju does that mean my possibility is valid?
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@Noel Lo – Yes, I think so. A lot of the wordings in this problem are somewhat ambiguous, though. Maybe the author thought along the same lines as X X while creating this, and actually meant that there have to be as many knaves as knights.
The way I think about it, though, you are perfectly correct in your logic.
If there are 2 knaves, then P3 is a knave, and not P1
Person 1: The number of either knights or knaves is the multiple of another of the different kind.
Person 2: There are less knaves than knights here.
Person 3: There is an odd number of knaves.
Person 4: The number of knaves is not a perfect square.
Person 5: The product of two numbers of knights and knaves is a perfect cube.
Person 6: The product of two numbers of knights and knaves is a perfect square.
Translated into probable sets as below, with (x,y) represent x Knight(s) and y Knave(s) :
P1 => anything but (6,0) or (0,6)
P2 => { (6,0) , (5,1) , (4,2) }
P3 => { (5,1) , (3,3) , (1,5) }
P4 => { (4,2) , (3,3) , (1,5) , (0,6) }
P5 => { (6,0) , (4,2) , (2,4) , (0,6) }
P6 => { (6,0) , (3,3) , (0,6) }
Done with the translation, let's pitch P1 against P5 against P6 in a truel. The three can't agree on one single thing between them, since #1 strictly believe in moderation but for the other two, even when they both are open to some degree of that moderation, theirs don't overlap on each others (one going for 'Pure Grey' while the other both 'Darker Grey' and 'Lighter Grey'). So we can conclude that we can always find a Knight and a Knave among P1, P5 & P6. This fact that reject the possibilities of both (6,0) and (0,6) found us our first Knight, P1.
Latest translation after some clean-up adjustment and modification looks like this :
P1 => anything but (6,0) or (0,6) ==> Knight
P2 => { (5,1) , (4,2) }
P3 => { (5,1) , (3,3) , (1,5) }
P4 => { (4,2) , (3,3) , (1,5) }
P5 => { (4,2) , (2,4) }
P6 => { (3,3) }
Just by looking at the translation, we saw that among the other unidentified islanders, "at most, only 3 out of the 5 can and will agree on one same thing(s)", either on the even split of the numbers with (3,3) or another with (4,2) unity.
Why are we even looking at the "maximal unity" here? It's because we are answering a question about "the least possible number of knaves there", and as there is only one consistent truth may be upheld at one time, we are using this fact to solve it by complementary method.
Now we know that no matter what, AT LEAST 2/6 will disagree with the majority (our P1 the Knight is agreeable to anything other the the extreme cases), so we can safely claim that "at least there are 2 Knaves among the group of 6 island people".
Full set of answers :
1) 2 Knaves
Knights = P1, P2, P4 & P5
Knaves = P3 & P6
2) 4 Knaves
Knights = P1 & P5
Knaves = P2, P3, P4 & P6
No matter what, P1 & P5 will always be Knights, P3 & P6 will always be Knaves while P2 & P4 can be either, but always stick together.
People may argue that 0/6 = 0 is a multiple of 6 (any numbers for that matter), and that still wouldn't change the fact that P1 will always be our first Knight to be found, since there IS a solution to this particular puzzle and P1 could not have lied with his broadly generic universal truth answer.
If a truel is too much to handle, then we can just pitch P3 against P5 in a duel.
Translated into probable sets as below, with (x,y) represent x Knight(s) and y Knave(s) :
P1 => anything but (6,0) or (0,6)
P2 => { (6,0) , (5,1) , (4,2) }
P3 => { (5,1) , (3,3) , (1,5) }
P4 => { (4,2) , (3,3) , (1,5) , (0,6) }
P5 => { (6,0) , (4,2) , (2,4) , (0,6) }
P6 => { (6,0) , (3,3) , (0,6) }
DUEL BETWEEN P3 & P5
P3 => { (5,1) , (3,3) , (1,5) }
P5 => { (6,0) , (4,2) , (2,4) , (0,6) }
These two are saying polar opposite things. One must be a Knight and the other knave. Group of 6 from the same kind possibilities is rejected by this finding. P1 is our first Knight.
If P1 is a knight,then there are 3 knights and 3 knaves.Then P2 is a knave,P3 is a knight,P4 is a knight,P5 is a knave,P6 is a knight. This case,there are 4 knights,so P1 must be a knave.P6 must be a knave,too.
If P5 is a knave,thenthere must be at least 5 knaves.If there are 6 knaves,then P4 is right,so it's impossible.If there are 5 knaves,then P3 and P4 are right,so it's impossible.So,P5 is a knight.There are 2 or 4 knaves.
If there are 4 knaves,then only P5 is telling the truth,so there are 2 knaves,P1 and P6.