Conventional Riddles

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Magic Square

3 x 3 square, all triples (across, down, and diagonal) must sum to 15. Use 1-9 once each.


One solution:
| 4 | 9 | 2 |
| 3 | 5 | 7 |
| 8 | 1 | 6 |

Mutilated Chessboard Problem

Two tiles, each from opposite corners of a chess board, are removed from the board. Is it possible for 31 dominoes, each covering 2 full tiles, to completely cover the remaining tiles on the board? [Solve without the most obvious method]

Solution No. Given the design of a chess board, each domino would have to cover both a light and dark tile, but the tiles at opposing corners share the same color.

Mutilated Cube Problem

Suppose you have 27 cubes stuck together, forming one larger (3 x 3 x 3) cube. What is the minimum number of straight cuts needed in order to separate all cubes from each other? After each cut, the pieces may be rearranged so that the next cut might separate more cubes. Prove your answer.

Solution The minimum number of cuts is 6. Proof: Consider the middle cube. There is no way to reveal any two of its faces in a single cut. And then it is trivial to see that it can indeed be done in 6 cuts.

Some Knights

How many knights can you place on a chessboard such that none of them can attack another?

Solution A knight's movement pattern necessitates that it lands on a square of the opposite color from the one it starts on. So the optimal placement would be 32 knights, one for each black (white) square on the board.

Two Ropes

You have two ropes, each of which will burn at varying speeds throughout their lengths, but by the end have burned for exactly 1 hour. How can you measure 45 minutes?

Solution Burn rope A at both ends and rope B at one. As soon as A is completely gone, light the other end of rope B.

Hard Boiled

You have a 7-minute hourglass and an 11-minute hourglass, how can you boil an egg in exactly 15 minutes?

Solution Start both hourglasses as you start boiling the egg. After the 7-minute hourglass runs out, turn it to start it again. Four minutes later, when the 11-minute hourglass runs out, turn the 7-minute hourglass again. Wait for the 7-minute hourglass to run out, which will take another four minutes and get you to exactly 15 minutes of boiling time.

The Great Divide

Four people arrive at a river with a narrow bridge that can only hold two people at a time. It’s nighttime and they have one torch that has to be used when crossing the bridge. Person A can cross the bridge in one minute, B in two minutes, C in five minutes, and D in eight minutes. When two people cross the bridge together, they must move at the slower person’s pace. Can they all get across the bridge in 15 minutes or less?

Solution Yes, 15: 1&2, 1, 5&8, 2, 1&2

A Very Contrived Situation

You buy a mansion. Right when you go inside, a robber jumps in through a window. He sees you and asks you to tell him which of the 3 switches in front of him turns on the attic light. If you lie to him, he will kill you. He lets you walk up the stairs once, but only once. Because you just bought the house, how will you know which switch turns on the attic light bulb?

Solution You turn the 1st switch on and wait for 10 minutes. Then, you turn the 1st switch off and the 2nd switch on. You immediately run up to the attic. If the light is on, the 2nd switch is the one that turns the light on. If the light is off, you feel the bulb. If the bulb is hot, it's the 1st switch. If the bulb is not hot, it's the 3rd switch.

Cheryl’s Birthday

Albert and Bernard just become friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates:


Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard doesn’t know too.
Bernard: At first I don’t know when Cheryl’s birthday is, but I know now.
Albert: Then I also know when Cheryl’s birthday is.

So when is Cheryl’s birthday?

Solution It’s July 16th. When Albert says his first line, it tells Bernard that the month is not May or June. These are the only months with a unique number, so for Albert to know Bernard doesn’t know right off the bat, it must be the case that Albert knows it’t in neither of those months. We know that narrowing it down to that extent is sufficient for Bernard to determine the birthday, which tells Albert it couldn’t have been the 14th. That information is sufficient for Albert to determine the birthday, because Albert knows the month. It would not be sufficient if the month was August.

Reservior Golfers

Four golfers named Mr. Black, Mr. White, Mr. Brown and Mr. Blue were competing in a tournament. The caddy didn’t know their names, so he asked them. One of them, Mr. Brown, told a lie.

The 1st golfer said “The 2nd Golfer is Mr. Black.”
The 2nd golfer said “I am not Mr. Blue!”
The 3rd golfer said “Mr. White? That’s the 4th golfer.”
The 4th golfer remained silent.

Which is which?

Solution In order: White, Black, Brown, Blue

Sweet Boxes

There are three opaque boxes. One with mints in it, another with aniseed sweets and in the third a mixture of the two. They are all labeled incorrectly. What is the minimum number of sweets you can take to work out which box is which?

Solution Just 1, from the box labeled mix.

The Gold Scale

There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?

Solution Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Use a triangle to easily calculate your expected weight sum should all the coins be fake, then .howevermanygrams over that value you are is how many pieces you put on the scale from that bag.

Billiard Balls

This one’s a wee bit ridiculous You have 12 billiard balls, one is heavier or lighter than the rest. You can use a justice-y scale up to 3 times to determine which ball is the odd one, and whether it’s heavier or lighter.

Solution The strat for the first weigh is to do 4&4, if they’re equal then you’ve narrowed it down to 4 and it’s trivial, if not equal, keep track of which are (possibly) heavier and lighter. For the second weigh put 2 of the lighter and 1 of the 4 you didn’t weigh (but know must be normal) on one side, and the other 2 lighter and a heavier on the other side. From there there’s a couple possibilities but with the knowledge you get from the first 2 weighs the options for the 3rd weigh become trivial ish. (For the 2nd, they obviously don’t have to be the lighter balls, just have to use 2 on each side of the balls that were on the same side for the 1st weigh, and note whether they’re heavier or lighter).

The Emperor’s New Game

You are a prisoner sentenced to death. The Emperor offers you a chance to live by playing a simple game. He gives you 50 black marbles, 50 white marbles and 2 empty bowls. He then says, ‘Divide these 100 marbles into these 2 bowls. You can divide them any way you like as long as you use all the marbles. Then I will blindfold you and mix the bowls around. You then can choose one bowl and remove ONE marble. If the marble is WHITE you will live, but if the marble is BLACK… you will die.’ How do you divide the marbles up so that you have the greatest probability of choosing a WHITE marble?

Solution Place 1 white marble in one bowl, and place the rest of the marbles in the other bowl (49 whites, and 50 blacks). This way you begin with a 50/50 chance of choosing the bowl with just one white marble, therefore life! BUT even if you choose the other bowl, you still have ALMOST a 50/50 chance at picking one of the 49 white marbles.

Island Fire

On island trapped by mountains, totally forest, starts on fire in west, wind blows it east until it burn everything on island, how survive?

Solution Take a lil burn, bring to unburn, burn, stand.

Two Guards

There are two paths forward, one will lead you where you want to go, the other will lead you to certain death. In front of each path stands a guard, one always lies, and the other always tells the truth. You don’t know which guard is which, nor which path is which. In one question, what can you ask one of the guards to find out which path will where you where you wish to go?

Solution Ask the guard "If I asked the other guard which path they guard, what would they say?"

Spot the Dog

Spot is a dog. Every dog has one alpha (also a dog), and no two dogs have the same alpha. But Spot isn’t anybody’s alpha. :( For any two dogs, there is an assigned referee in case they get into a fight and an assigned marriage counselor in case they get married. The dogs have set up the following rules for deciding who will be the referees and counselors for who: If Spot fights with any dog, the other dog gets to be the referee. The referee of a fight between (dog 1) and (dog 2’s alpha) has to be the alpha of the referee of a fight between dogs 1 and 2. Spot has to be his own marriage counselor, no matter who he marries. The marriage counselor for (dog 1) and (dog 2’s alpha) has to referee any fight between dog 1 and the marriage counselor for dog 1 and dog 2. Who’s the marriage counselor for Spot’s alpha and the referee of a fight between Spot’s alpha and himself (Spot’s alpha)?

Solution Spot’s Alpha’s Alpha.

The Sum and Product Puzzle

xx and yy are two different whole numbers greater than 11. Their sum is not greater than 100100, and yy is greater than xx. Person SS knows the sum x+yx + y, and person PP knows the product x×yx \times y. Both SS and PP know all the information in this paragraph. The following conversation occurs:

SS: “PP does not know xx and yy
PP: “Now I know xx and yy
SS: “Now I know xx and yy

What are xx and yy?


Below is the script I wrote to solve this puzzle. It doubles as a detailed description of the reasoning behind the solution. Alternately, read more about the puzzle here.

import math

X_MIN   = 2
X_MAX   = 49
Y_MIN   = 3
Y_MAX   = 98
SUM_MAX = 100

# Return all valid pairs of numbers that sum to n
def sum_pairs(n):
    pairs = []
    for a in range(X_MIN, X_MAX + 1):
        b = n - a
        if b <= a:
        pairs.append((a, b))
    return pairs

# Return all valid pairs of factors of n
def factor_pairs(n):
    lower_bound = X_MIN
    upper_bound = math.floor(math.sqrt(n))
    pairs = []
    for p in range(lower_bound, upper_bound + 1):
        if n % p == 0:
            q = n // p
            if q > Y_MAX:
            if p + q > SUM_MAX:
            if p == q:
            pairs.append((p, q))
    return pairs

# S: “P does not know X and Y”
# This implies that among all of the possible pairs of numbers summing to S's number,
# none of those pairs multiply to a number with just one valid pair of factors.
possible_pairs = set()
for sum in range(SUM_MAX + 1):
    pairs = sum_pairs(sum)
    valid = True
    for (x, y) in pairs:
        if (len(factor_pairs(x * y)) == 1):
            valid = False
    if valid:

# P: “Now I know X and Y”
# This implies that P's number did have multiple valid factor pairs, 
# but only one pair appeared in the list resulting from S's first statement.
possible_products = set()
duplicate_products = set()
for (x, y) in possible_pairs:
    if x * y in possible_products:
        duplicate_products.add(x * y)
        possible_products.add(x * y)

unique_product_pairs = set()
for (x, y) in possible_pairs:
    if x * y not in duplicate_products:
        unique_product_pairs.add((x, y))

# S: “Now I know X and Y”
# This implies that S's number did have multiple valid sum pairs, 
# but all of those pairs except for one were ruled out by their presence
# in the set of duplicate product pairs we obtained from P's statement.
possible_sums = set()
duplicate_sums = set()
for (x, y) in unique_product_pairs:
    if x + y in possible_sums:
        duplicate_sums.add(x + y)
        possible_sums.add(x + y)

unique_sum_pairs = set()
for (x, y) in unique_product_pairs:
    if x + y not in duplicate_sums:
        unique_sum_pairs.add((x, y))

# The fact that we are able to find X and Y implies that the set of pairs
# has been reduced to one. This is the solution.

The Impossible Chessboard Problem

The jailer will take you into a private cell. In the cell will be a chessboard and a jar containing 64 coins.

The jailer will take the coins, one-by-one, and place a coin on each square on the board. He will place the coins randomly on the board. Some coins will be heads, and some tails (or maybe they will be all heads, or all tails; you have no idea. It’s all at the jailers whim. He may elect to look and choose to make a pattern himself, he may toss them placing them the way they land, he might look at them as he places them, he might not …). All you can do is watch.

Once all the coins have been laid out, the jailer will point to one of the squares on the board, indicating the one with the key under it (invisible from your position). The jailer will then allow you to turn over any one coin on the board. This is the only change you are allowed to make to the jailers initial layout. You will then be lead out of the room, and the jailer will bring in your friend, who will examine the board of coins and decide under which square the key is. They have only one chance to guess correctly, in which case you’re both set free.

The jailer explains all these rules, to both you and your friend, beforehand and then gives you time to confer with each other to devise a strategy for which coin to flip.


Actually I haven’t solved this yet, but the solution should be here.