Friends Tom, Dick and Harry are each either a knight, always telling the truth, or a liar, always lying. One day they went to the donut shop and bought six donuts.
Tom claimed to have gotten only one donut, while Dick claimed Tom had two. Harry said Tom had more than three.
All three agreed that Dick had two donuts.
If each of the three got at least one donut and no donuts were shared or divided, how many donuts did each friend get?
We denote the three friends Tom, Dick and Harry respectively by T, D and H. Let the respective number of donuts possessed by T, D and H be x, y and z.
We note that the three friends agreed upon the number of donuts possessed by Dick, but all three differed on the matter regarding how many donuts Tom had. This is possible only when each of T, D and H are liars. Since each of T, D and H must possess at least one donut, we observe that the three friends together had falsely stated that T either had less than three donuts or T had more than three donuts.
Therefore, T must have possessed precisely three donuts, giving x =3, so that y+z = 6-3 = 3.
However, in terms of the false agreement between the three friends that H had 2 donuts, it follows that y!= 2. Since, none of x, y and z can be zero, the only solution to y+z = 3 is (y,z) = (1,2)
Consequently,
Tom possessed three donuts
Dick possessed one donut, and:
Harry possessed two donuts.
Edited on November 6, 2007, 2:08 pm
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