Web Puzzles

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.