Hello everyone! Congratulations to everyone who attempted the RMO 2024. As you might know, we had an amazing livesolve of the paper with Archit, Adhitya, Abel and Kanav which you can check out here . We also have question wise video solutions to all the problems, thanks to Nanda, Om and Shreya! We had a lot of people interested in solutions for the KV/JNV paper, which is what this blog post will be about. Without further ado, let's get started! Problem 1: Find all positive integers $x,y$ such that $202x+4x^2=y^2$. Solution: Notice that $y>2x$. Let $y=2x+k$ for some integer $k>0$. Thus, the given equation reduces to $$202x=4xk+k^2\implies x=\frac{k^2}{202-4k}\cdots (1)$$ This tells us that $202-4k|k^2,$ or that $101-2k|2k^2\implies 101-2k|101k$. However, since 101 is a prime, $\gcd(101-2k,\,101)=1\implies 101-2k|k$ or that $101-2k|2k\implies 101-2k|101\implies k=50$. Substituting in $(1)$, we get that $x$ must be $$\frac{50^2}{202-4(50)}=50\cdot 25=1250.$$ Thus, $$y=2x+5
In this post, I will present three graph theory problems in increasing difficulty, each with a common theme that one would determine a property of an edge in a complete graph through repeated iterations, and seek to achieve a greater objective. ESPR Summer Program Application: Alice and Bob play the following game on a $K_n$ ($n\ge 3$): initially all edges are uncolored, and each turn, Alice chooses an uncolored edge then Bob chooses to color it red or blue. The game ends when any vertex is adjacent to $n-1$ red edges, or when every edge is colored; Bob wins if and only if both condition holds at that time. Devise a winning strategy for Bob. This is more of a warm-up to the post, since it has a different flavor from the other two problems, and isn't as demanding in terms of experience with combinatorics. However, do note that when this problem was first presented, applicants did not know the winner ahead of time; it would be difficult to believe that Bob can hold such a strong