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Problem 151 Solution

1. Pure and sane
2. Pure and insane
3. Applied and sane
4. Applied and insane

The first step is to determine what each kind of mathematician would say about themselves:

1. Pure and sane: "I am pure and sane."
2. Pure and insane: "I am applied and sane."
3. Applied and sane: "I am pure and insane."
4. Applied and insane: "I am applied and insane."

Note that the applied and insane mathematician is lying about a lie so is actually speaking the truth.

Next consider the first statement by Alice, "I am insane." Based on the four groups there are two that will say they are insane:

1. The applied and sane.
2. The applied and insane.

By the same logic we can determine that Bob is sane, Charlie is insane, and Dorothy is pure.

Now we know one of the two characteristics of each mathematician. Next jump to the last statement by Dorothy, "Charlie is sane." We already know that Dorothy is pure so she actually believes that Charlie is sane. However we also know that Charlie is infact insane so Dorothy's belief must be incorrect. Since she is not deliberately telling a lie her belief is incorrect, making her insane. So Dorothy is pure and insane.

Let review what we know thus far:

1. Alice: applied.
2. Bob: sane.
3. Charlie: insane.
4. Dorothy: pure and insane.

Next look at what Bob says about Dorothy, "Dorothy is insane." This is a true statement and since Bob is already sane we can conclude that he is also pure. Now we have:

1. Alice: applied.
2. Bob: sane and pure.
3. Charlie: insane.
4. Dorothy: pure and insane.

Next consider what Charlie says about Bob, "Bob is applied." This is not true since Bob is pure. We already know Charlie is insane thus he must also be pure. If he were applied he would be lying about a lie, thus telling the truth. Now we have:

1. Alice: applied.
2. Bob: sane and pure.
3. Charlie: pure and insane.
4. Dorothy: pure and insane.

Finally consider Alice's second statement, "Charlie is pure." Charlie is pure thus Alice is making a true statement. However Alice is applied, thus she thinks she is telling a lie. So Alice is lying about an incorrect belief, making a true statement. Since Alice's belief is incorrect she must be insane. Thus:

1. Alice: applied and insane.
2. Bob: pure and sane.
3. Charlie: pure and insane.
4. Dorothy: pure and insane.

I'd like to thank Mathematics and Informatics Quartly 9:4 Dec 1999, page 162 for this problem.

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