JotD / QotD Ελληνική Λίστα Κουίζ (QotD)

Θέμα: Re: Κάτι για να σκεφτεστε ...

(nil): Akis Karnouskos (akis(@)
Ημερομηνία: Πεμ 30 Απρ 1998 - 17:57:28 EEST

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.

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)

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.

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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