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LMAO Revenge

Continuing the tradition of past years, our seniors at the Indian IMO camp(an unofficial one happened this year) once again conducted LMAO, essentially ELMO but Indian. Sadly, only those who were in the unofficial IMOTC conducted by Pranav, Atul, Sunaina, Gunjan and others could participate in that. We all were super excited for the problems but I ended up not really trying the problems because of school things and stuff yet I solved problem 1 or so did I think. Problem 1:  There is a   grid of real numbers. In a move, you can pick any real number  ,  and any row or column and replace every entry   in it with  .  Is it possible to reach any grid from any other by a finite sequence of such moves? It turned out that I fakesolved and oh my god I was so disgusted, no way this proof could be false and then when I was asked Atul, it turns out that even my answer was wrong and he didn't even read the proof, this made me even more angry and guess wha...
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Functional Equations 101

Let's get to the math:  Let there be two sets X and Y. A function  from X to Y denoted as f:X→Y is assigning a value in Y for every element in X. We say that X is the domain of the function f and Y is the range.  A function f:X→y is said to be injective if f(x)=f(x′)⟹x=x′ To put it in a more abstract way, if there is some a∈Y then there is at most one b∈X such that f(b)=a holds true.  A function is said to be surjective when for any a∈Y there is at least one b∈X such that f(b)=a holds true.  A function is bijective if for every a∈Y there is exactly one b∈X such that f(b)=x. Bijective functions are basically functions which are both injective and surjective.  Bonus: A function f:X→X is known as an involution if f(f(x))=x∀x∈X  As an exercise, the readers should try to prove that every function th...
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RMO 2024: Discussing Solutions

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+4x2=y2. Solution:  Notice that y>2x. Let y=2x+k for some integer k>0. Thus, the given equation reduces to 202x=4xk+k2⟹x=k2202−4k⋯(1) This tells us that 202−4k|k2, or that 101−2k|2k2⟹101−2k|101k. However, since 101 is a prime, gcd(101−2k,101)=1⟹101−2k|k or that 101−2k|2k⟹101−2k|101⟹k=50. Substituting in (1), we get that x must be $$\frac{50^2}{202-4(50)}=50\cdot 25=12...
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