My conclusion is that Jones is the truth teller.
Suppose Robinson is the truth teller.
Robinson:
1. True I owe no money to anybody.
2. True Jones is a Dutchman.
3. True I always tell the truth.
4. True Two of Jones’s statements are true, and two are false.
then:
Jones: (must be 2T and 2F per Robinson 4)
1. True I owe no money to Smith.
2. False Robinson owes me $7. per Robinson 1
3. False I am British. per Robinson 2
4. True All Smith’s statements are untrue.
and then:
Smith: must be all True
1. False Jones owes me $10. Contradiction with Jones 1
2. False Robinson owes me $5. Contradiction with Robinson 1
3. True All Robinson’s statements are true.
4. False All Jones’s statements are untrue, because Jones must be 2T and 2F.
So it's not Robinson.
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Suppose Smith is the truth teller.
Smith: all True by assumption.
Smith:
1. True Jones owes me $10.
2. True Robinson owes me $5. Inconsistent with Robinson 1.
3. True All Robinson’s statements are true.
4. True All Jones’s statements are untrue.
then:
Robinson: per Smith 3 all must be True
1. True I owe no money to anybody. Inconsistent with Smith 2.
2. True Jones is a Dutchman.
3. True I always tell the truth.
4. True Two of Jones’s statements are true, and two are false.
So it's not Smith
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Suppose Jones is the truth teller.
Jones: all True by assumption.
1. True I owe no money to Smith.
2. True Robinson owes me $7.
3. True I am British.
4. True All Smith’s statements are untrue.
then:
Smith: must be all False according to truth teller Jones 4
1. False Jones owes me $10.
2. False Robinson owes me $5.
(Robinson owes Smith either 0 or some number but not 5.)
3. False All Robinson’s statements are true. Robinson has at least one lie.
4. False All Jones’s statements are untrue. (but we're assuming they're all True)
and then:
Robinson:
1. False I owe no money to anybody. per Jones 2
2. False Jones is a Dutchman. per Jones 3
3. False I always tell the truth.
(because Robinsons 1 and 2 are False)
4. False Two of Jones’s statements are true, and two are false.
So all of Jones statements are True
All of Smith's and Robinson's statements are False.
p&p solution
