**(nil):** Akis Karnouskos (*akis(@)ceid.upatras.gr*)

**Ημερομηνία:** Πεμ 30 Απρ 1998 - 17:57:28 EEST

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1) 29

Each person paid $9, totalling $27. The manager has $25 and the bellboy

$2. The bellboy's $2 should be added to the manager's $25 or subtracted

from the tenants' $27, not added to the tenants' $27.

2) BOXES

The problem cannot be solved with the information given.

The sign on box A says "The sign on box B is true and the gold is in box

A". The sign on box B says "The sign on box A is false and the gold is

in box A". The following argument can be made: If the statement on box A

is true, then the statement on box B is true, since that is what the

statement on box A says. But the statement on box B states that the

statement on box A is false, which contradicts the original assumption.

Therefore, the statement on box A must be false. This implies that

either the statement on box B is false or that the gold is in box B. If

the statement on box B is false, then either the statement on box A is

true (which it cannot be) or the gold is in box B. Either way, the gold

is in box B.

However, there is a hidden assumption in this argument: namely, that

each statement must be either true or false. This assumption leads to

paradoxes, for example, consider the statement: "This statement is

false." If it is true, it is false; if it is false, it is true. The only

way out of the paradox is to deny that the statement is either true or

false and label it meaningless instead. Both of the statements on the

boxes are therefore meaningless and nothing can be concluded from them.

In general, statements about the truth of other statements lead to

contradictions. Tarski invented metalanguages to avoid this problem. To

avoid paradox, a statement about the truth of a statement in a language

must be made in the metalanguage of the language.

Common sense dictates that this problem cannot be solved with the

information given. After all, how can we deduce which box contains the

gold simply by reading statements written on the outside of the box?

Suppose we deduce that the gold is in box B by whatever line of

reasoning we choose. What is to stop us from simply putting the gold in

box A, regardless of what we deduced? (cf. Smullyan, "What Is the Name

of This Book?", Prentice-Hall, 1978, #70)

3) CAMELS

The wiseman tells them to switch camels.

4) Children

First, determine all the ways that three ages can multiply together to

get 72: (quite a feat for the bartender)

72 1 1

36 2 1

24 3 1

18 4 1

18 2 2

12 6 1

12 3 2

9 4 2

9 8 1

8 3 3

6 6 2

6 4 3

As the man says, that's not enough information; there are many

possibilities. So the bartender tells him where to find the sum of the

ages--the man now knows the sum even though we don't. Yet he still

insists that there isn't enough info. This must mean that there are two

permutations with the same sum; otherwise the man could have easily

deduced the ages.

The only pair of permutations with the same sum are 8 3 3 and 6 6 2,

which both add up to 14 (the bar's address). Now the bartender mentions

his "youngest"--telling us that there is one child who is younger than

the other two. This is impossible with 8 3 3--there are two 3 year olds.

Therefore the ages of the children are 6, 6, and 2.

Pedants have objected that the problem is insoluble because there could

be a youngest between two three year olds (even twins are not born

exactly at the same time). However, the word "age" is frequently used to

denote the number of years since birth. For example, I am the same age

as my wife, even though technically she is a few months older than I am.

And using the word "youngest" to mean "of lesser age" is also in keeping

with common parlance. So I think the solution is fine as stated.

In the sum-13 variant, the possibilities are:

11 1 1

10 2 1

9 3 1

9 2 2

8 4 1

8 3 2

7 5 1

7 4 2

7 3 3

6 6 1

6 5 2

6 4 3

The two that remain are 9 2 2 and 6 6 1 (both products equal 36). The

final bit of info (oldest child) indicates that there is only one child

with the highest age. This cancels out the 6 6 1 combination, leaving

the childern with ages of 9, 2, and 2.

5) CONDOMS

Use both condoms on the first woman. Take off the outer condom (turning

it inside-out in the process) and set it aside. Use the inner condom

alone on the second woman. Put the outer condom back on. Use it on the

third woman.

First man uses both condoms. Take off the outer condom (do NOT reverse

it) and have second man use it. First man takes off the inner condom

(turning it inside-out). Third man puts on this condom, followed by

second man's condom.

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