Deduction 6 - Puzzle Prime

Ten Lanterns

Section: Brain Teasers Category: Deduction

Difficulty: Advanced

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.

SOLUTION

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.

Unknown Author December 27, 2020

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Buffalo Buffalo Buffalo

Section: Brain Teasers Category: Deduction

Difficulty: Medium Tags: Words

The sentence below is grammatically correct. Can you explain it?

Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo.

SOLUTION

The sentence says that buffalo (animals) from Buffalo (city, US), which are buffaloed (intimidated) by Buffalo (city, US) buffalo (animals), themselves buffalo (intimidate) buffalo (animals) from Buffalo (city, US).

Unknown Author December 24, 2020

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Eight Queens Puzzle

Section: Brain Teasers Category: Deduction

Difficulty: Advanced

Place 8 queens on a chessboard, so that no two of them attack each other. For an extra challenge, make sure that no three of them lie on a straight line.

SOLUTION

The original puzzle has 12 unique solutions, up to rotation and symmetry. With the additional restriction imposed, there is only one solution.

Max Bezzel November 27, 2020

5 Comments

Einstein’s Puzzle

Section: Brain Teasers Category: Deduction

Difficulty: Hard

There are 5 houses and each of them has a different color. Their respective owners have different heritages, drink different types of beverages, smoke different brands of cigarettes, and look after different types of pets. It is known that:

The Brit lives in the red house.
The Swede keeps dogs as pets.
The Dane drinks tea.
Looking from in front, the green house is just to the left of the white house.
The green house’s owner drinks coffee.
The person who smokes Pall Malls raises birds.
The owner of the yellow house smokes Dunhill.
The man living in the center house drinks milk.
The Norwegian lives in the leftmost house.
The man who smokes Blends lives next to the one who keeps cats.
The man who keeps a horse lives next to the man who smokes Dunhill.
The owner who smokes Bluemasters also drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
The man who smokes Blends has a neighbor who drinks water.

The question is, who owns the pet fish?

SOLUTION

The German owns the pet fish.

Since the Norwegian lives in the leftmost house (9) and the house next to him is blue (14), the second house must be blue. Since the green house is on the left of the white house (4), the person living in the center house drinks milk (8), and the green house’s owner drinks coffee (5), the fourth house must be green and the fifth one must be white. Since the Brit lives in the red house (1) and the Norwegian lives in the leftmost house (9), the leftmost house must be yellow and the center house must be red. Therefore, the colors of the houses are: YELLOW, BLUE, RED, GREEN, WHITE.

Since the Norwegian from the yellow house smokes Dunhill (7), the man from the blue house must keep a horse (11). The person smoking Blends cannot be in the red house, because this would imply that the person in the green house keeps cats and the Swede keeps dogs in the white house (2, 10). However, in this case the Dane must be drinking tea in the blue house (3) and the person smoking Blends does not have a neighbor drinking water (5), which is a contradiction (15). Also, the person smoking Blends cannot be in the green house, because this would imply that the person in the white house drinks water (15), the Dane lives in the blue house (3), and the German and the Swede live in the last two houses. Since the German smokes Prince (13) and the Swede keeps dogs (2), there is nobody who could smoke Bluemaster and drink beer (12). The person smoking Blends cannot be in the white house either, because this would imply that the person in the green house drinks water (15), when in fact he drinks coffee (5).

Therefore, the person smoking Blends must be in the blue house, and then the German and the Swede must live in the last two houses (2, 13). Since the person who smokes Bluemasters drinks beer (12), this must be the Swede with his dogs in the white house (2). The only option for the person who smokes Pall Mall and raising birds (6) is the red house. Then the Norwegian must keep cats (10) and the German is left with the pet fish in the green house.

Albert Einstein October 24, 2020

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King Arthur and the Knights

Section: Brain Teasers Category: Deduction

Difficulty: Medium

King Arthur and his eleven honorable knights must sit on a round-table. In how many ways can you arrange the group, if no honorable knight can sit between two older honorable knights?

SOLUTION

The answer is 1024 ways, up to rotation around the table. To see this, note that the youngest honorable knight must sit right next to King Arthur – there are two possible places for him. Then, the second-youngest knight must sit right next to this group of two. Once again, there are two possible places for him. Continuing like this, we see that for all honorable knights, except for the oldest one, there are two possible spots on the table. Multiplying two to the power of ten out, we get 1024.

Unknown Author August 6, 2020

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Non-Transitive Dice

Section: Brain Teasers Category: Deduction

Difficulty: Advanced

This is a non-transitive dice set, i.e. every dice in it is weaker than some other dice. Can you design a non-transitive set with only 3 dice?

Remark: “Weaker” means that it loses more often than it wins.

SOLUTION

The simplest solution is given by:

2, 2, 4, 4, 9, 9;
1, 1, 6, 6, 8, 8;
3, 3, 5, 5, 7, 7.

Another solution is given by the so-called “Miwin’s dice”. They are as follows:

1, 1, 3, 5, 5, 6;
2, 3, 3, 4, 4, 5;
1, 2, 2, 4, 6, 6.

Unknown Author May 9, 2020

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Socks in the Drawer

Section: Brain Teasers Category: Deduction

Difficulty: Easy

You have a drawer with 10 pairs of black socks and 10 pairs of white socks. How many times do you need to blindly reach inside the drawer and take out a sock, so that you get a matching pair?

SOLUTION

Only 3 times. Once you have two socks of the same color, they already form a matching pair.

Unknown Author May 3, 2020

2 Comments

Rich and Poor

Section: Brain Teasers Category: Deduction

Difficulty: Easy

Who is richer – the richest among the poor or the poorest among the rich?

SOLUTION

The poorest among the rich is richer than the richest among the poor. That is because any rich person is by default richer than any poor person.

Unknown Author January 19, 2020

1 Comment

Prank the Professors

Section: Brain Teasers Category: Deduction

Difficulty: Medium

Three professors fell asleep under a tree. At some point a prankster passed by and painted their faces with black dye. When the professors woke up, each of them saw the others’ faces and started laughing at them. After a while though, they stopped laughing, realizing that their own faces were painted as well. How did they deduce that?

SOLUTION

Let us denote the professors with A, B and C. The thought process of A would go like this: “If my face is not painted, then B will see that C is laughing at him and will realize immediately that he is being pranked. However, B was laughing for a while and therefore my I must being pranked as well.”. Then A will stop laughing and the same will happen with the other two professors B and C.

Unknown Author December 22, 2019

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Guess the Fruits

Section: Brain Teasers Category: Deduction

Difficulty: Medium

You are given 3 boxes – one labeled “Apples”, one labeled “Bananas”, and one labeled “Apples and Bananas”. You are told that the labels on the boxes have been completely mismatched, i.e. none of the three labels is put on its correct box. How can you open just one box and pick a random fruit from it, so that after seeing the fruit, you can guess correctly the contents of every box out of the three?

SOLUTION

Open the box labeled “Apples and Bananas”. If you pick a banana from it, then the box labeled “Bananas” will contain apples, and therefore the box labeled “Apples” will contain apples and bananas. Similarly, if you pick an apple from it, then the box labeled “Apples” will contain bananas, and therefore the box labeled “Bananas” will contain apples and bananas”.

Unknown Author December 4, 2019

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