Web Puzzles

Island X Puzzle Solution

Assume that A is the Truther. If so, then B is the Liar as A's statement asserts. If so, B's second statement is false, so C is the Alternator. This implies that C's first statement is false as also his second statement that A is the Liar, so that C is the Liar which is a contradiction, so that A cannot be the Truther.

Assume that B is the Truther. If so, his first statement is a direct contradiction, implying amongst other things, that B cannot be the Truther.

Assume that C is the Truther. If so, then A is the Liar in conformity with his second statement, so the remaining member B must be the Alternator. This checks out since as a Alternator, B's first statement is true while in his second statement he falsely identified C as the Liar. Both of A's statement are then clearly false, so this establishes the veracity of both the statements of C.

Consequently, (A, B, C) = (Liar, Alternator, Truther)

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.

Switching Logic Puzzle Solution

Turn switch #1 ON. After about five minutes or so, turn switch #1 OFF and turn switch #2 ON.
Then go upstairs and check on the bulbs.

The one ON is obviously #2. The other ones are OFF, but one of them should be very hot by having been ON for five minutes. That's #1, and the remaining bulb is #3.

The Diagram Puzzle Solution

Joe counted how many times each of the 7 LEDs lit up for all 10 digits...

 _
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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.

Carpet Layer Puzzle Solution

The only measurement that needs to be taken is the distance between opposite angles of the rectangular dance floor (figure on the left). That distance then will be squared (figure on the right) and doubled to get the sum of the areas of the 4 squares.

Socks And Shops Puzzle Solution

For each of the twenty pairs of socks, they unclipped it and gave one sock each.

Strawberry Ice Cream Puzzle Solution

First, determine all the ways that three ages can multiply together to get 72:

• 72 1 1 (quite a feat for the bartender)
• 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.

The Package Puzzle Solution

With the object in the box, G locks it with one of his own locks. He then sends the box to H. H attaches his own lock in addition to G's and then sends it back. G removes his lock and then sends it to H again.

Napoleon's Star Puzzle Solution

To be able to place nine coins, it is necessary to make the 3rd point of each step equal to the start point of the previous step. For example:

a-g; i-a; c-i; f-c; e-f; h-e; b-h; j-b; d-j.


 With such a simple solution, it's hard to believe that Napoleon stayed in charge for so long.

Barrels 'O' Fun Puzzle Solution

There are multiple ways of solving this. One way is given below, and it's probably the fastest one. Each set of 3 numbers separated by hyphens is the amount of wine (in litres) in the 3 barrels after each "pouring operation". The 3 barrels are always in the same order: 12, 7, and 5 litres.

• 12-0-0
• 5-7-0
• 5-2-5
• 10-2-0
• 10-0-2
• 3-7-2
• 3-4-5
• 8-4-0
• 8-0-4
• 1-7-4
• 1-6-5
• 6-6-0.

Faulty Batches Puzzle Solution

They had to weigh 1 coin from the 1st batch, 2 from the 2nd, 4 from the 3rd, 8 from the 4th, and 16 from the 5th one.

If all coins weighed 10 grams as they should, the scale would display 310 grams ((1 + 2 + 4 + 8 + 16) * 10). However, since one batch has 9 grams coins, and another 11 grams coins, then the total weight of this combination of coins will be:

Total Weight	Number of 9g coins	Number of 11g coins
311	1	2
313	1	4
317	1	8
325	1	16
312	2	4
316	2	8
324	2	16
314	4	8
322	4	16
318	8	16
309	2	1
307	4	1
303	8	1
295	16	1
308	4	2
304	8	2
296	16	2
306	8	4
298	16	4
302	16	8

After seeing the solution to this puzzle, it is clear that it would be a lot easier to simply use the scales up to 5 times rather than go through all this, but where is the fun in that?

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.