Four mathematicians have the following conversation: Alice: I am insane. Bob: I am pure. Charlie: I am applied. Dorothy: I am sane. Alice: Charlie is pure. Bob: Dorothy is insane. Charlie: Bob is applied. Dorothy: Charlie is sane. You are also given that: Pure mathematicians tell the truth about their beliefs. Applied mathematicians lie about their beliefs. Sane mathematicians\' beliefs are correct. Insane mathematicians beliefs are incorrect. Describe the four mathematicians. Of three men, one man always tells die truth, one always tells lies yes or no randomly. Each man knows which man is which. you much no questions to determine who is who. If you ask the same question one person you must count it as a question used for each person a questions should you ask? Tom is from die census bureau and greets Mary a, her door. They conversation: Solution 7.56 Notice that a pure mathematician will always describe his or herself as sane (regardless of his or her actual sanity). Meanwhile an applied mathematician will always describe his or herself as insane. Therefore Alice is applied and Dorothy is pure. In a similar fashion we can reason that Charlie is insane, and Bob is sane. Now Dorothy’s statement that “Charlie is sane” is incorrect, and since we know Dorothy is pure we reason that she is also insane. Then Bob was correct when he stated “Dorothy is insane”, and since we know him to be sane he must also be pure. Thus Charlie was incorrect when he stated ”Bob is applied”, and since we know Charlie to be insane he must also be pure. Finally that means Alice was correct when she stated “Charlie is pure”, and since we already know she is applied, she must also be insane. 7.57) There are six orders they could be in: TRL, TLR, RTL, RLT, LTR, and LRT. Start by asking the 1st man a question. No matter what your question is, if the first man is the random one, he could answer either way. So RTL and RLT will be in both groups. The trick is to ask a question that will allow you to know where the random one isn\'t. So you want TRL and LRT to answer one way and TLR and LTR to answer the other. So I would ask person 1: Consider one of the other two who does not answer randomly. If I were to ask him if person 2 is the random one, would he answer yes? If I get a yes answer, I know the order is RTL, RLT, TLR, or LTR. If I get a no answer, I know the order is RTL, RLT, TRL, or LRT. If I get a yes, I know that person 2 is not the random one, so I will ask him if he is the random one. If he says yes, he is the liar (so RLT or TLR). For my third question, I will ask person 2 if person 1 is the random one. Yes means TLR, no means RLT. If he says no to question 2, he is the truth teller (so RTL or LTR). Again, I will ask him if person 1 is random. Yes means RTL and no means LTR. If I get a no, I know that person 3 is not the random one, so I will ask him if he is the random one. If he says yes, he is the liar (so RTL or TRL). For my third question, I will ask person th.

1. Four mathematicians have the following conversation: Alice: I am insane. Bob: I am pure.
Charlie: I am applied. Dorothy: I am sane. Alice: Charlie is pure. Bob: Dorothy is insane.
Charlie: Bob is applied. Dorothy: Charlie is sane. You are also given that: Pure mathematicians
tell the truth about their beliefs. Applied mathematicians lie about their beliefs. Sane
mathematicians' beliefs are correct. Insane mathematicians beliefs are incorrect. Describe the
four mathematicians. Of three men, one man always tells die truth, one always tells lies yes or
no randomly. Each man knows which man is which. you much no questions to determine who is
who. If you ask the same question one person you must count it as a question used for each
person a questions should you ask? Tom is from die census bureau and greets Mary a, her door.
They conversation:
Solution
7.56
Notice that a pure mathematician will always describe his or herself as sane (regardless of his or
her actual sanity). Meanwhile an applied mathematician will always describe his or herself as
insane. Therefore Alice is applied and Dorothy is pure. In a similar fashion we can reason that
Charlie is insane, and Bob is sane. Now Dorothy’s statement that “Charlie is sane” is incorrect,
and since we know Dorothy is pure we reason that she is also insane. Then Bob was correct
when he stated “Dorothy is insane”, and since we know him to be sane he must also be pure.
Thus Charlie was incorrect when he stated ”Bob is applied”, and since we know Charlie to be
insane he must also be pure. Finally that means Alice was correct when she stated “Charlie is
pure”, and since we already know she is applied, she must also be insane.
7.57)
There are six orders they could be in: TRL, TLR, RTL, RLT, LTR, and LRT. Start by asking the
1st man a question. No matter what your question is, if the first man is the random one, he could
answer either way. So RTL and RLT will be in both groups. The trick is to ask a question that
will allow you to know where the random one isn't. So you want TRL and LRT to answer one
way and TLR and LTR to answer the other.
So I would ask person 1: Consider one of the other two who does not answer randomly. If I
were to ask him if person 2 is the random one, would he answer yes? If I get a yes answer, I
know the order is RTL, RLT, TLR, or LTR. If I get a no answer, I know the order is RTL, RLT,
TRL, or LRT.
If I get a yes, I know that person 2 is not the random one, so I will ask him if he is the random

2. one. If he says yes, he is the liar (so RLT or TLR). For my third question, I will ask person 2 if
person 1 is the random one. Yes means TLR, no means RLT. If he says no to question 2, he is
the truth teller (so RTL or LTR). Again, I will ask him if person 1 is random. Yes means RTL
and no means LTR.
If I get a no, I know that person 3 is not the random one, so I will ask him if he is the random
one. If he says yes, he is the liar (so RTL or TRL). For my third question, I will ask person three
if person 1 is the random one. Yes means TRL and no means RTL. If he says no, he is the truth
teller (so RLT or LRT). Again, I will ask him if person 1 is random. Yes means RLT and no
means LRT.
7.58)
From the statement "the product of their ages is 36" the possibilities of the three individual ages
are:
1, 1, 36
1, 2, 18
1, 3, 12
1, 4, 9
1, 6, 6
2, 2, 9
2, 3, 6
3, 3, 4
From the statement "the sum of their ages is the same as my house number," it is possible to
eliminate all but two possibilities. The sums of these answers we can eliminate are "unique"
and if any of them were the house number, then Tom would have then known the ages of the
kids!
For example if Mary's house number were 38 he would know that the ages must be 1, 1, and 36!
If her house number were 21, he would know that the ages must be 1, 2, and 18.
If her house number were 10, Tom would know that the ages of her kids must be 3, 3, 4 etc.
So, because of this, these six possibilities can all be eliminated:
1, 1, 36 = 38
1, 2, 18 = 21
1, 3, 12 = 16
1, 4, 9 = 14
2, 3, 6 = 11
3, 3, 4 = 10
The only two remaining possibilities are 1, 6, 6, and 2, 2, 9. (Mary's house number is therefore

3. 13 which is why at this point Tom says, "I still don't know their ages.")
After the clue "the younger two are twins" you can obviously eliminate 1, 6, 6. The only
remaining possibility is then 2, 2, 9!
Of course if Mary had said "the older two are twins" then the answer would indeed be 1, 6, 6!!
1, 1, 36
1, 2, 18
1, 3, 12
1, 4, 9
1, 6, 6
2, 2, 9
2, 3, 6
3, 3, 4