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Knavish Arithmetic (Posted on 2019-05-25) Difficulty: 5 of 5
The participants in our little logical mystery, are two pairs, each consisting of one knight and one knave.

The first pair, made up S1 and S2, are told sum of two integers (x + y). While the second pair, made up of P1 and P2, are told the product (xy).

At the outset, none of the participants know the identity of the knave in the other pair, although they are aware that each group is mixed.

Further, all participants have been told that 1 < x < y < 100.

Interaction between the four participants takes place as follows. In the initial phase, each participant writes a statement, initially hiding it from all other participants. The four statements are then revealed to all participants simultaneously

Initial phase:

•S1 wrote: “I deduce that 64 < xy < 196”

•S2 wrote: “It is impossible for P1 and P2 to deduce x and y from xy at this point”

•P1 wrote: “It is impossible for S1 and S2 to deduce x and y from x + y at this point”

•P2 wrote: “I deduce that x+y = 33”

Following these disclosures, a sequence of remarks are made by the participants in the following order:

Follow-up conversation:

•S1 says “It is impossible to determine which P is the knave from the above statements alone”

•P1 says “Now I know x and y”

•S2 says “Now I know x and y”

•P2 says “P1 and S2 are knaves”

Identify the two knaves and determine x and y, if you can!

See The Solution Submitted by FrankM    
Rating: 4.6667 (3 votes)

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Some Thoughts Knaves Comment 2 of 2 |
According to Raymond Smullyan, knaves always lie. On Perplexus, knaves alternate between truths and lies. It seems that knaves always lie in this problem, because if knaves alternate truths and lies, then we get a contradiction.

If P2 is a knight, then P2's first statement is true. If P2 is a knave, then P2's second statement is false since there are only two knaves. Then, P2's first statement is true. Therefore, P2 can deduce that x+y=33. Then, either x=2, y=31, xy=62, or x=11, y=22, xy=242. Then, S1's first statement is false, so S1 is a knave. Since P2 could deduce x and y, then S2 is a knave. Therefore, both S1 and S2 are knaves. That is a contradiction. Therefore, knaves in this problem always lie. You should have called them liars.


  Posted by Math Man on 2019-06-21 11:04:42
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