If you want hints or discuss a solution to one of the problems, you can find my information in the 'contact me' tab. New problems added!
No particularly fancy math is required to solve any of these problems.

Problem 1

$$j^j = $$ Where \(j = \sqrt{-1} \)

Problem 2 (New!)

Is this true?

$$e^j = (e^j)^{\displaystyle\frac{2\pi}{2\pi}} = (e^{j2\pi})^{\displaystyle\frac{1}{2\pi}} = 1^{\displaystyle\frac{1}{2\pi}} = 1 $$

Problem 3

$$\sum_{n=1}^\infty \displaystyle\frac{n^2}{2^n} = $$

Problem 4

Express
$$\sqrt[\leftroot{9}\uproot{3}8]{2207 - \displaystyle\frac{1}{2207 - \displaystyle\frac{1}{2207 - \displaystyle\frac{1}{2207 - ...}}}}$$ as \(\displaystyle\dfrac{a + b\sqrt{c}}{d}\) where \(a,b,c,d \in \mathbb{R}\)

Problem 5

There are \(3\) jars each with an incorrect label. Each jar contains \(100\) beans. One jar has \(100\) white beans, another has \(100\) black beans, and the third has \(50\) white and \(50\) black beans. WHat is the least amount of beans required to draw and from which jar to know with absolute certainty which jar is which?

Problem 6 (New!)

\(20\) prisoners are given a shot at their redemption. They are told that on the next day they will be lined up such that they could only see all the people in front of them. Each prisoner will be given a hat, either black or white (assume an infinite supply of hats), and each prisoner must guess the color hat he or she is wearing. The last person goes first, and everyone can hear his response and the consequence (being let free or killed). Is there a way for the prisoners to collaborate, to save more than \(10\) people?

Problem 7

Three men decide to go fishing on a remote island. It gets too late for them to go home so they decide to sleep on the island and divide what was caught the next day. The first guy can't fall asleep so he decides to count the fish, sees there is one too many to divide evenly by 3 so he throws one fish out, takes his share and goes home. The second guy wakes up a little later and wants to go home. He doesn't realize that one has left, he counts the fish and realizes that there is one too many to divide evenly by 3, so he throws one out, takes his share and goes home. In the morning the third guy wakes up and wants to go home. He doesn't realize hat both his buddies left him, so he counts the fish and realizes that there is one too many to divide evenly by 3, so he throws one fish out, takes his share and goes home. How many fish did the three guys start off with in order for this to be possible? Is your answer unique?