Cut out the black shape and then fold it perfectly into a cube, without overlapping.
SOLUTION
If the size of each of the small edges of the shape is equal to 1cm, then its area is equal to 30cm². Thus, the cube that is formed by the shape must have an edge of length √5cm. If we set the center of the black shape to be the center of one of the cube’s faces, then the four closest vertices must lie on the boundary of the shape, √2.5cm away. This determines uniquely the folding of the shape which is shown on the simulation below.
Three friends, A, B and C, want to find out what their average salary is without disclosing their own salaries to the others. How can they do it using only verbal communication?
SOLUTION
A tells B some number, then B adds his salary to it and tells the result to C, then C adds his salary and tells the result to A. Now A subtracts the number he told B in the beginning, adds his own salary and divides by 3. Repeat the same procedure with B and C starting first.
Help Mrs. Vampire get to her sweet home safely, avoiding all the light coming from outside. Of course, she can close the doors if she gets close enough to them.
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.
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.
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:
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