We at Mystery City Games love history, culture and secrets. Therefore, we have combined those things into Amsterdam's first historical puzzle treasure hunt. Come try it out!
August 20, 1940
Leon Trotsky, an enemy of Josef Stalin, had been in exile for 11 years. On this day, a Soviet assassin entered Trotsky’s house in Mexico City and hit him in the head with an ice axe. He died the next day.
Here is a group of suspects who were rounded up. Each wrong suspect has one different feature from the correct suspect. Who killed Trotsky?
The killer of Leon Trotsky was Ramón Mercader.
If all plinks are plonks and some plunks are plinks, which of these statements must be true?
Remark: “Some” means more than 0.
The first statement says that the set of plonks contains the set of plinks, and the second statement says that there is at least one plunk-plink. Therefore, that plunk-plink must also be a plonk, and the second statement is true.
The first and the third statements, however, do not need to be true. Indeed, it is possible that there is a plink that is not a plunk, or that all plinks are plunks.
The solution to this puzzle is unique, but you don’t need this information in order to find it.
Remark: The answer to question 20. is (E).
The answers are:
In the three murder cases below, you can read the testimonies of all suspects. For each case, find who the killer is, knowing that no 2 people are in the same row or column, and that the killer was alone in a room with the victim.
The solutions are shown below.
There are three playing cards in a row. There is a two to the right of a king. There is a diamond to the left of a spade. There is an ace to the left of a heart. There is a heart to the left of a spade. Identify the three cards.
The cards are an Ace of Diamonds, a King of Hearts, and a Two of Spades.
Cheryl’s birthday is one of 10 possible dates:
May 15
May 16
May 19
June 17
June 18
July 14
July 16
August 14
August 15
August 17
Cheryl tells the month to Albert and the day to Bernard.
Albert: I don’t know the birthday, but I know Bernard doesn’t know either.
Bernard: I didn’t know at first, but now I do know.
Albert: Now I also know Cheryl’s birthday.
When is Cheryl’s birthday?
If Albert knows that Bernard doesn’t know when the birthday is, then the birthday can’t be on May 19 or June 18. Also, Albert must know that the birthday can’t be on these dates, so May and June are completely ruled out.
If Bernard can deduce when the birthday is after Albert’s comment, then the birthday can’t be on 14th. The remaining possibilities are July 16, August 15, and August 17.
Finally, if Albert figures out when the birthday is after Bernard’s comment, then the date must be July 16.
In a small village, there are 100 married couples living. Everyone in the village lives by the following two rules:
One day a traveler comes to the village and finds out that every man has cheated at least once on his wife. When he leaves, without being specific, he announces in front of everybody that at least one infidelity has occurred. What will happen in the next 100 days in the village?
Let us first see what will happen if there are N married couples in the village and K husbands have cheated, where K=1 or 2.
If K = 1, then on the first day the cheating husband would get killed and nobody else will die. If K = 2, then on the first day nobody will get killed. During the second day, however, both women would think like this: “If my husband didn’t cheat on me, then the other woman would have immediately realized that she was being cheated on and would have killed her husband on the first day. This did not happen and therefore my husband has cheated on me.” Then both men will get killed on the second day.
Now assume that if there are N couples on the island and K husbands have cheated, then all K cheaters will get killed on day K. Let us examine what will happen if there are N + 1 couples on the island and L husbands have cheated.
Every woman would think like this: “If I assume that my husband didn’t cheat on me, then the behavior of the remaining N couples will not be influenced by my family’s presence on the island.” Therefore, she has to wait and see when and how many men will get killed in the village. After L days pass however and nobody gets killed, every woman who has been cheated on will realize that her assumption is wrong and will kill her husband on the next day. Therefore, if there are N + 1 couples on the island, again all L cheating husbands will get killed on day L.
Applying this inductive logic consecutively for 3 couples, 4 couples, 5 couples, etc., we see that when there are 100 married couples on the island, all men will get killed on day 100.
Warning: this puzzle involves mature themes that are inappropriate for younger audiences. If you are not an adult, please skip this puzzle.
Mary is 21 years older than her son. After 6 years, she will be 5 times older than him. Where is the father?
Let M be the age of the mother and S be the age of the son. We have M = S + 21 and M + 6 = 5(S + 6). We solve the system and get S= -3/4, i.e. minus 9 months. Therefore right now the son just got conceived and the father is with the mother.
At Creepy Beasts Inc., three of the most dreaded animals, a tiger, a wolf, and a bear, sat in their boardroom in silence while they awaited their boss. Then, Mr. Tiger broke the silence.
“Isn’t it odd that our three surnames are the same as our three species, yet none of our surnames matches our own species?”
The wolf replied, “Yeah, but does anyone care?”
They sat in silence again…
Can you figure out the surname of each animal?
Since the wolf replied to Mr. Tiger, his surname can be neither Tiger nor Wolf. Therefore, the wolf’s surname is Mr. Bear. Subsequently, Mr. Tiger must be a bear, and finally, Mr. Wolf must be a tiger.
You have ten lanterns, five of which are working, and five of which are broken. You are allowed to choose any two lanterns and make a test that tells you whether there is a broken lantern among them or not. How many tests do you need until you find a lantern you know for sure is working?
Remark: If the test detects that there are broken lanterns, it does not tell you which ones and how many (one or two) they are.
You need 6 tests:
(1, 2) → (3, 4) → (5, 6) → (7, 8) → (7, 9) → (8, 9)
If at least one of these tests is positive, then you have found two working lanterns.
It all of these tests are negative, then lantern #10 must be working. Indeed, since at least one lantern in each of the pairs (1, 2), (3, 4), (5, 6) is not working. Therefore, there are at least 2 working lanterns among #7, #8, #9, #10. If #10 is not working, then at least one of the pairs (7, 8), (7, 9), or (8, 9) must yield a positive test, which is a contradiction.
Please confirm you want to block this member.
You will no longer be able to:
Please note: This action will also remove this member from your connections and send a report to the site admin. Please allow a few minutes for this process to complete.
