Web Puzzles

The Greatest Show On Earth Puzzle Solution

There are 69 bipeds and 43 four-legged animals. If all animals were four-legged, the officer would have counted 448 legs, not 310. Obviously 138 legs are missing, hence 69 subjects are bipeds.

It's worth pointing out that, within the bipeds, there are 13 stilt birds, which include flamingos and storks. The cook must have counted them as they were standing on one leg only, so whilst he is not crazy, he almost drove the customs officer mad.

A Thinking Man Puzzle Solution

The symbol ¿ means the difference between the number made up of the all digits of the operation, and the mirror of this last number. i.e,
5 ¿ 2 = 52 - 25 = 27
6 ¿ 3 = 63 - 36 = 27
8 ¿ 4 = 84 - 48 = 36

An alternative solution for this symbol (submitted by Alfa Chan... many thanks!) is simply the difference between the two numbers multiplied by 9. i.e,
5 ¿ 2 = (5 - 2) × 9 = 27
6 ¿ 3 = (6 - 3) × 9 = 27
8 ¿ 4 = (8 - 4) × 9 = 36

Another alternative solution submitted by Mickey Kawick... thanks!! We have x ¿ y; If x is odd, then the result is 5x + y, otherwise it's 5x - y. i.e,
5 ¿ 2 = 5 × 5 + 2 = 27 (5 is odd, so we add the 2)
6 ¿ 3 = 6 × 5 - 3 = 27 (6 is even, so we subtract the 3)
8 ¿ 4 = 8 × 5 - 4 = 36 (8 is even, so we subtract the 4)

The symbol ¤ means the difference between the two numbers multiplied by the larger of the two numbers. i.e,
5 ¤ 2 = (5 - 2) * 5 = 3 * 5 = 15
6 ¤ 4 = (6 - 4) * 6 = 2 * 6 = 12
3 ¤ 8 = (8 - 3) * 8 = 5 * 8 = 40

An alternative solution for this symbol, as submitted by Jeff Schall (many thanks!), is the difference between the square of the bigger of the two numbers and their product. i.e,
5 ¤ 2 = (5 ^ 2) - (5 * 2) = 25 - 10 = 15
6 ¤ 4 = (6 ^ 2) - (6 * 4) = 36 - 24 = 12
3 ¤ 8 = (8 ^ 2) - (3 * 8) = 64 - 24 = 40

Three Divine Comedians Puzzle Solution

Dante figured that first of all, he had to find out which of the three monsters was the lunatic one. Let's call the three monsters A, B, C.

In order to do so, he asked one of them (let's say monster A for this example) a question like: "If I asked a question to monster B, would I stand a greater chance of obtaining a truth than if I asked the same question to monster C?"

The possible combinations (where + means the monster that tells the truth, - means the monster that lies, x means the lunatic monster) are:

Answer given	Monster A	Monster B	Monster C
No	+	-	x
Yes	+	x	-
No	-	+	x
Yes	-	x	+
Yes/No	x	+	-
Yes/No	x	-	+

If the answer was a "Yes", then it was safe to say that x couldn't have been monster C; on the other hand, if the answer was a "No", then x couldn't possibly have been Monster B.

Once that Dante obtained this information, he used his second question to find out whether the monster that is definately NOT the lunatic one was + or -, and the easiest way of achieved this was by asking a question with an obvious answer such as, "Is 2 an even number?" After finding out whether the monster is the liar or the sincere one, the third question was used to resolve the other two monsters; a question like "Is monster A the lunatic one" did the trick.

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.

Three Camels Puzzle Solution

The Great Sage added his own camel to the other seventeen. He then gave 9 camels (one half of 18) to the eldest of the three, 6 camels (one third) to Ali, and 2 camels (one 9th) to Ismail. Then took his own camel back and sat in front of the tent, thanking God for His generosity.

Another way to explain the solution, perhaps in a more mathematical manner, is the following, courtesy of Gopalakrishnan Thirumurthy.
The three sons are assigned their shares: 1/2, 1/3, and 1/9. The sum of their shares is 1/2 + 1/3 + 1/9 = 17/18. Out of 18 camels, 17 of them are left by their father. So,

• #1 gets 1/2 of 18 = 9
• #2 gets 1/3 of 18 = 6
• #3 gets 1/9 of 18 = 2

9 + 6 + 2 = 17.

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).

Faulty Batch Puzzle Solution

The secretary placed on the scale 1 coin from the first batch, 2 from the second, and so on until he put 8 from the eighth batch.

If all coins weighed 10 grams each, then the weight displayed on the scale should have been 360 grams ((1 + 2 + ... + 8) × 10). But, since one batch of coins weighs less, the difference between 360 grams and the weight displayed on the scale should point us to the faulty batch. For example, if the faulty batch was the fifth one, then the total weight displayed on the scale would be 355 grams. Or if it was the seventh batch, the weight would have been 353 grams, ie 7 grams less than the theoretical total weight of 360 grams.

An 'optimisation' on this solution is to omit the 8 coins from the eighth batch. In this case, the maximum weight of the coins would be 280 grams, and if it equals 280, then the eighth batch is the faulty one. Thanks to Denis Borris for this observation.

By using the same logic, one could omit the coins from any one of the other batches, instead of the eighth one. For example, if we omit the fourth batch, we'll be left with a theoretical 320 grams and, if it is indeed the total weight, then we will know that the fourth batch was the faulty one. Thanks to Glen Parnell for noticing this.

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.

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.

Ivanovic had to turn around two posters: the first one (Lenin) and the third one (the Red Star).

All that Nikita Proskoijev said was that all the posters representing Lenin show Hammer And Sickle on their opposite face. Therefore it is needed to check the back of the first poster and the front of the third one, to make sure that the Red Star wasn't linked with Lenin's face, cause if it was, then Proskoijev's statement would have been false.

Checking the front of the fourth poster (which is what Ivanovic did, that's why he's now a lumberjack near Jakutsk) is useless; if the 4th poster shows Lenin face as its front, it would just confirm what Proskoijev stated; but then, if the 4th poster showed Marx, this would not have falsified Proskoijev's statement, because he said that Lenin is linked to Hammer And Sickle, while he didn't state that Hammer And Sickle are linked to Lenin.

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.