Web Puzzles

A Special Old Guy Puzzle Solution

Firstly, we can say that the date of birth (DOB) for the special old guy (SOG) is:
Equation {a}:
DOB(SOG) = x^2 – x

Similarly, for the son of the special old guy (or SOSOG for short):

Equation {b}:
DOB(SOSOG) = y^3 – y

SOG's age when SOSOG was born was:
Equation {c}:
Age(SOG) = DOB(SOSOG) – DOB(SOG) = (y^3 – y) – (x^2 – x)

There are unlimited solutions to equation {c} so we need some assumptions and limits.

We know that SOG must have been between, say 10 years old and 100 years old when SOSOG was born:

Equation {d}:
10 < Age(SOG) < 100

We also know that both SOG and SOSOG were born some time since the year 500 AD:
Equation {e}:
500 < DOB(SOSOG) < 2004.

Based on equation {b} and {e} we can clearly see that there are only 5 solutions for y. They are y = 8, 9, 10, 11 or 12. Any other solutions for y are in breach of equation {e}.
For each of these possible solutions for y there is only a limited number of solutions for x that comply with {d}. They are:

 x     y
--------
21     8
22     8
26     9
27     9
31    10
36    11
41    12

Any other solutions for x and y are in breach of equation {d}.
SOG's age when his son was born can be calculated for each of these possible solutions by using equation {c} as follows:

 x      y     Age(SOG)
----------------------
21      8           84
22      8           42
26      9           70
27      9           18
31     10           60
36     11           60
41     12           76

So there are 7 different solutions, which is why Buzz said he couldn't solve for x and y.
Now comes the lateral part of the puzzle: Alvin informs Buzz that "...he was your age when his son was born". Of course, Buzz knows his own age. He should therefore be able to pick the correct solution from the list of 7 possible solutions shown above. However, he can't. That means that Buzz (and the special old guy) must be 60 because there are multiple solutions for an age of 60, whereas any other age would yield a unique solution. Any other age and Buzz would be able to solve.
A 60 year old father – quite a special old guy.

How Many Monks? Puzzle Solution

The way they work this out and the reason it happens on the same day is as follows. It is important to note that not only are the monks intelligent but they all know all their fellow monks are as well.
Start with one monk:

He knows he is the only one who can have the disease so he kills himself on
day 1.

Then two monks:
They know that either one of them has the disease or both do. If monk1
doesn't see the mark of the disease on monk2 then he will realise that he
must have it so will kill himself. If he sees the mark then he knows that
monk2 will following the same reasoning and kill himself. If the next day
the other monk is still alive then he realises that the other monk must have
seen the mark on him and so they both have the disease and he must kill
himself. Both follow the same reasoning and kill themselves on day 2.

Three monks:
If only one of the monks has the disease he will see no mark on the other
two and so diagnose himself. He will kill himself on day one. If two monks
have the disease then they will each see one monk with the mark and one
without. When they see each other again the next day they will deduce that
the monk they see with the mark would only not have killed himself if he
could see someone else with the mark. They know it is not the third monk so
it must be them. Both diseased monks follow the same reasoning and kill
themselves on day two. If all three monks have the disease then they are all
in the same position as the healthy monk in looking on in the previous
example. Each of them can see two diseased men. When they haven't both
killed themselves on day two there can be only one reason - the viewing monk
must have the mark as well. All taking the same reasoning they all kill
themselves on day three.

And so on...
N monks all with the disease will all kill themselves on day N.

Weighing an Elephant Puzzle Solution

Load the elephant onto a boat large enough to carry it. The boat will sink slightly, and you mark the level of the water on the side of the boat. Then you offload the elephant and fill the boat with bags until the boat sinks to the level marked. The bags can be individually weighed using beam scales and the weight of the elephant is the sum of the weight of the bags.

This puzzle is slightly cunning in that it the geographic location of the city is a small clue.

Cheers To Statistics Puzzle Solution

Three artist would do the trick. This is because twenty-four artists would paint twenty-four models in two and a half hours. Since the available time increases eight-fold (2.5 * 8 = 20), it is possible to reduce the number of painters by the same number of times (24 / 8 = 3).

Orbiting Logic Puzzle Solution

During the next orbit, the sleepy but correct answer came forth from the astronaut: 2 of clubs, 5 of hearts, 7 of diamonds, 9 of spades, and 10 of hearts.

Following from (g) - no two cards are the same rank - and (b), the strongest combination (ie highest ranks) possible is 10, 9, 7, 6, 4, which adds up to 36. If we write down all combinations of 2 numbers that have a difference of 9, and a maximum sum of 36, we'll have:

• 2, 11
• 3, 12
• 4, 13
• 5, 14
• 6, 15
• 7, 16
• 8, 17
• 9, 18
• 10,19
• 11, 20
• 12, 21
• 13, 22

The number of odd ranks in the 5-card combination could be 1 or 3 (not 2 and not 4 because their sum would be an even number, and a difference of 9 between two even numbers is inexistent). If there was only 1 odd rank, then there would be 4 even ranks, and the weakest combination of evens would be 2, 4, 6, 8 which add up to 20: the difference, 9, would result in a single number, 11, which does not exist in the deck of cards (the Royals were excluded). Therefore we have 3 odd ranks. The sum of the weakest 3 odd ranks (excluding the Aces) is 15, ie 3 + 5 + 7. So, from the combinations above, we can exclude all combinations that contain an odd number less than 15. We are left with

• 6, 15
• 8, 17
• 10, 19
• 12, 21

The first one (6, 15) must be excluded because the 3 odd numbers (3, 5, 7) would be joined by the only 2 even numbers which would add up to 6 (2,4), and therefore we would have a sequence of 5 consecutive numbers, which doesn't match the constraint set by (b). We are left with 3 combinations:

• 8, 17 (and the 5 ranks would be 2, 3, 5, 6, 9)
• 10, 19 (and the 5 ranks would be 3, 4, 6, 7, 9)
• 12, 21 (and the 5 ranks would be 2, 5, 7, 9, 10)

But (d) states that the sum of red cards is twice of the sum of black cards, so the first 2 combinations must be excluded, because it's not possible to find 2 sets of numbers, one of which is twice the other; therefore the 5 ranks can only be 2, 5, 7, 9, 10. The red cards must be 5, 7, 10.

Following (e), rank 2 must be clubs and 10 must be hearts, so rank 9 must be spades; following (f), rank 7 has got to be diamonds and rank 5 must be hearts.

Three Skullcaps Puzzle Solution

At first, any explorer could have guessed the colour of his own skullcap only if the other two wore green skullcaps. Unfortunately, the first explorer admits to not being able to work it out, and is killed.

Then, with two people left it is possible for either explorer to know if he wears a red skullcap only if the other wears green. In that case, he could reason, "The other person wears green, so if I also have a green skullcap, then the first man would have deduced that he was wearing a red skullcap, since there are only two green caps. Therefore my skullcap is, without doubt, red." Of course, this is not the case. Stupidly, the second explorer admits he does not know and is killed as punishment.

After seeing that the other two could not deduce their colours, and believing in their deductive capabilities, the third prisoner was then sure he was wearing a red skullcap.

Mizar Puzzle Solution:

11 seconds. The 5 seconds needed to signal six o'clock are the 5 silent intermissions between rings. At twelve, the 12 rings are interleaved by 11 silent intermissions, which need 11 seconds to be executed.