Problem List
These are the problems posted on the year 2017. Click to view!

01: Unstacking Game (Dec 29, 2017)  Suppose you have a stack of \(n\) blocks. A move is defined as taking \(k \left(\leq p-1\right)\) blocks from a stack of \(p\left(\geq 2\right)\) blocks and splitting the original stack into two stacks, each with blocks \(k\) and \(p-k\). For each move you get \(k(p-k)\) points.
  For instance, you get \(3 \times 5=15\) points for moving \(3\) blocks from a stack of \(8\) blocks.
 The player applies a move until there are \(n\) stacks, each with only \(1\) block left.

 No matter what strategy you use, show that the total sum of points \(S(n)\) is always the same, and find \(S(n)\).

Solution (By Strong Induction) Let \[S(n) = \frac{n(n-1)}{2}\] For \(n = 2\), you can only move \(1\) block, and get \(S(2) = \frac{2\times 1}{2} = 1\) points. Formula for \(S(n)\) is true for this case.
Suppose the formula for \(S(i)\) is true for \(1 \leq i \leq n\).
When there are \(n+1\) blocks, take \(k \leq n+ 1\) blocks from the stack.
You will get \(k(n+1-k)\) points for this move.
From the induction hypothesis, from the remaining two stacks, you get \(S(k)+ S(n+1-k)\) points. Thus, you get a total of \[S(n+1)=k(n+1-k)+S(k)+S(n+1-k)\] points. Using the formula for \(S(n)\), \[S(n+1)=k(n+1-k)+\frac{k(k-1)}{2}+\frac{(n+1-k)(n-k)}{2} \] \[\therefore S(n+1)=\frac{(n+1)n}{2}\] Therefore the formula is true for \(n+1\).
No matter what strategy you use, the total sum of points will be equal to \(S(n)=\frac{n(n-1)}{2}\).

02: Tangent of Sum of Arctangents (Aug 22, 2017)  Let \(t\) be defined as follows, \[\Large t=\sum_{i=1}^{\infty} \arctan{\frac{1}{2i^2}}\]
  Find \(\tan t\).