Puzzle Archives - Puzzle Prime

Six Glasses

Section: Brain Teasers Category: Insight

Difficulty: Medium

Six identical glasses are placed in a row on the table – first three filled with water, and then three empty ones. Can you move just one glass, so that empty and full glasses alternate?

SOLUTION

Take the second full glass, pour all the water into the second empty glass, and then put it back in its place.

Unknown Author March 1, 2024

2 Comments

Celebrity Names 3

Section: Casual Puzzles Category: Rebus

Difficulty: Easy

What celebrity names do these cave paintings represent?

SOLUTION

The answers are:

(Mice – Ice) + Eye + Yell + Fast + Bender = Michael Fassbender
Dam + Sander = Adam Sandler
Ant on a Knee + Hop + Kins = Anthony Hopkins

Mark Irish February 28, 2024

1 Comment

Three Inch Origami

Section: Brain Teasers Category: Insight

Difficulty: Advanced

What is the minimum number of folds you need in order to create a 3-inch measurement from an 8.5×11 inches paper?

SOLUTION

You need just 2 folds, as shown on the picture below.

Source:

Puzzling StackExchange

DEEM February 25, 2024

0 Comments

XKCD Crossword

Section: Brain Teasers Category: Deduction

Difficulty: Easy

Can you solve XKCD’s crossword puzzle?

SOLUTION

All letters in the crossword are “A”.

xkcd February 22, 2024

0 Comments

Protect the Treasure

Section: Brain Teasers Category: Mathematics

Difficulty: Advanced

Nine pirates have captured a treasure chest. In order to protect it, they decide to lock it using multiple locks and distribute several keys for each of these locks among them, so that the chest can be opened only by a majority of the pirates. What is the minimum number of keys each of the pirates should get?

SOLUTION

First, we show that for every four pirates, there exists a lock which cannot be opened only by them and can be opened by everyone else. We choose an arbitrary group of four pirates. If they can open every lock, then they can access the treasure without the need of a majority. If any of the remaining pirates cannot open that lock, then he, together with the initial group of four still cannot access the treasure. Thus, the claim is proved and to each group of four pirates we can assign a unique lock. These are $\binom{9}{4}=126$ locks in total. Finally, every pirate should get keys for $\binom{8}{4}=70$ of these locks, one for each group of four additional pirates he can be a group of.

Unknown Author February 19, 2024

2 Comments

How Many Times on Average?

Section: School Category: Quant Interview

Tags: Probability

A common type of questions which appears on Quant Interviews is:

“You have a sequence of random variables. How many times on average does a certain outcome appear in this sequence?”

Here are a few examples:

You are throwing a coin 100 times. How many times will you encounter the Heads-Heads-Tails in this sequence?
Five husbands and five wives have sat around a circle table in random order. What is the average number of spouses which are sitting next to each other?

While such problems can be solved using induction, there is another, more elegant approach. It is based on the following observation:

The average number of successful events is equal to the sum of the probabilities that each of these events is successful. The events do NOT need to be independent.

Written mathematically, this translates to:

\mathbb{E}(\|\{X_i = 1, 1 \leq i \leq n \}\|) = \mathbb{E}(\sum_{i=1}^{n} X_i) = \sum_{i=1}^{n} \mathbb{E}(X_i) = \sum_{i=1}^{n}\mathbb{P}(X_i =1),

where $X_i$ is a random variable, such that $X_i=1$ if the i-th event is successful $X_i = 0$ otherwise.

The first equation follows from the definition of $X_i$ .
The second equation follows from linearity of expectation.
The third equality is a basic property of expectation.

Let us see how to apply this technique to solve the coin question posed above.

We define a random variable $X_i$ which is equal to $1$ if and only if coin tosses $i, i+1, i+2$ are Tails-Tails-Heads respectively. We have:

\mathbb{E}(\|\{X_i=1, 1\leq i \leq 100\}\|) = \sum_{i=1}^{100}\mathbb{P}(X_i=1)

It is easy to see that for each $1\leq i \leq n, X_i = \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$ . Therefore, the answer to the problem is $100\times \frac{1}{8}=12.5$ .

To solve the second problem, we define a random variable $X_i$ which is equal to $1$ if and only a husband and his wife are sitting on spots $i$ and $i+1$ around the table. For each $i$ the probability of this happening is equal to $\frac{1}{9}$ . Indeed, no matter who sits on spot $i$ , the chance that their spouse sits on spot $i+1$ is 1:9. Therefore, the average number of spouses sitting next to each other is:

\mathbb{E}(\|\{X_i=1, 1\leq i \leq 10\}\|) = \sum_{i=1}^{10}\mathbb{P}(X_i=1)=10\times\frac{1}{9}=1\frac{1}{9}

As we can see, the presented technique is a simple, but very powerful tool. For extra practice, try to solve this fun puzzle from our blog on your own:

Fish Eat Fish

Puzzle February 16, 2024

0 Comments

Buried Up to Neck

Section: Brain Teasers Category: Deduction

Difficulty: Medium

Three friends, Adam, Bob, and Charlie are buried in the sand up to their necks, all facing West. Charlie can see both Adam and Bom, Bom can see only Adam, and Adam cannot see anyone. Black and white hats are placed on their heads. The three friends are told that there is at least one hat from each color, and then they are asked whether anyone can guess the color of their own hat.

After a few minutes, one of them answers. Who is that?

SOLUTION

If Adam and Bob had hats with identical colors, then Charlie would immediately be able to deduce that his hat has the opposite color. Charlie doesn’t do that, so Adam and Bob are able to figure out that their hats have opposite colors. Since Bob is the one who can see the color of Adam’s hat, he is the one that answers the question.

Unknown Author February 13, 2024

0 Comments

The Letter E

Section: Brain Teasers Category: Practical

Difficulty: Expert

Click the image to enlarge it and download it. Then print it, cut out the three pieces, and rearrange them so that you get the letter “E”.

SOLUTION

Rearrange the pieces, so that they form the shadow of the letter “E”, as shown below.

Unknown Author February 10, 2024

1 Comment

Fish Eat Fish

Section: Brain Teasers Category: Mathematics

Difficulty: Advanced Tags: Probability

A hundred fish are swimming along a stream at different velocities. If one fish catches up to another fish, it eats it and continues swimming. What is the expected number of fish that will survive?

SOLUTION

Notice that the N-th fish in the stream survives if and only if it is the fastest among the first N fish. The probability of this event happening is equal to 1/N. Since the expected number of fish that survive is equal to the sum of the survival probabilities for each of them, the answer is 1+1/2+1/3+…+1/N.

For more details on the last claim, consider reading our blog post “How Many Times on Average?”

Unknown Author February 7, 2024

0 Comments