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Showing posts from July, 2022

A phase I'll never forget...

 It was the night of 8th August 2021, when we had the closing ceremony of CAMP (A math camp organized by Ritam Nag, Aatman Supkar and Rohal Goyal). My eyes were a bit wet realizing that it is ending and I won't have those 3 - 4 hours of fun every night which I had the whole past week. Thinking that time about all the memories I made in the camp from joining the game nights to getting frank with my seniors, it was probably one of the best experiences I had in any of the math related thing I did. I was fascinated by the organization of this camp and was just wondering if I was good enough to ever do something like this in future, but I never really got enough courage or time to do anything like this until recently. Fast forwarding to the time after INMO results, me and many of my friends were very sad that IMOTC has been cancelled by HBCSE and we couldn't meet each other for another year. Well, jokingly in a meet around that time, Sunaina told me to ask Rohan if they can organize

Top 10 problems of the week!

This week was full Number Theory and algebra biased!! 😄 Do try all the problems first!! And if you guys get any nice solutions, do post in the comments section! Here are the walkthroughs of this week's top 5 Number Theory problems! 5th position (1999 JBMO P2):  For each nonnegative integer $n$ we define $A_n = 2^{3n}+3^{6n+2}+5^{6n+2}$. Find the greatest common divisor of the numbers $A_0,A_1,\ldots, A_{1999}$. Walkthrough:  It doesn't require a walkthrough, I wrote this here, cause it's a cute problem for the person who has just started Contest math :P a. What is $A_0$? b. Find out $A_1$. c. Show that $\boxed{7}$ is the required answer! 4th position (APMO, Evan Chen's orders modulo a prime handout):   Let $a,b,c$ be distinct integers. Given that $a | bc + b + c, b | ca + c + a$ and $c | ab + a + b$, prove that at least one of $a, b, c$ is not prime. Walkthrough:  Fully thanks to  MSE  ! (Also one should try MSE, it has helped me a lot, ofc it's more tilted to Coll

Whose Game?

 In this blog post, we will talk about Combinatorial games, how to quantify positions, and analyse games. Most of this is based on the beginning chapters of the comprehensive book Winning Ways for your Mathematical Plays by Berlekamp, Conway and Guy. What is a Game? You might be familiar with multiple 'games', all with greatly varied meanings. For example, a  game could refer to a wild animal hunted for sport (This post isn't about those!). We will talk about games that are defined by the following (restrictive) rules: There are just 2 players, often named Left and Right. There are several well-defined positions and a starting position. There are clearly defined rules specifying the moves each player can make from a position, and the resulting positions are called options. Left and Right move alternately in gameplay. In the normal play convention, the player unable to move loses. In misere  play, the last player to move loses. The rules ensure that the game ends and one p

Trust Issues

Hi, I am Debayu and this is my second blogpost !! We start off with the following problem.  Problem: Find the sum of all $b$ such that $$2n+1 \mid \binom{2n}{b} -1$$ for all $n$ such that $2n \geq b$. At first glance, this looks somewhat innocent and its probably some problem from some random computational contest, some may this can be what the IOQM 10 markers will look like this year. hmmm so what do we do here? Let's check what happens when $b$ is $2$. So we have that for all $n$, $$2n+1 \mid \binom{2n}{2}$$ $$\iff 2n+1 \mid n(2n-1)-1=2n^2-n-1$$ $$ \iff 2n+1 \mid |2n^2-n-1-n(2n+1)| = |-2n-1|$$ which is true so yay $2$ works. Now we might want to check $b=3$, this means $$2n+1 \mid \binom{2n}{3}$$ $$\iff  2n +1 \mid \frac{n(2n-1)(2n-2)}{3}$$ $$\iff 6n+3 \mid 4n^3-6n^2+2n$$ from here, it is easy to see that if $n$ is even, the divisibility wont hold so $3$ doesn't work. Similarly one can check that $4$ doesn't work either and one might guess that the only possible $b$ is $2

Algorithms, or Mathematics?!

Hi everyone! In my last blog, I spoke about a couple of very interesting algorithmic techniques and how they can be used to solve a variety of very interesting problems. This blog however, is going to be completely different.   When we’re young we begin by learning the steps to add – we’re given the rules and we must learn to repeat them – no questions asked. Why does carry forward work the way it does? Well, no questions asked. To some extent, it is important to know how exactly to do a certain set of things while first learning maths. We may not be able to get anywhere in a subject if we’re unable to learn a few basic rules and know how to use them. However, after a certain point it is important to bring in the spirit of mathematical thinking within each student too – something missing in almost every form of school math education. Mathematical miseducation is so common, we wouldn’t even see it. We practically expect a math class to look like repetition and memorisation of disjointed

Mobius Inversion

The word 'Mobius' would instantly remind the readers of the Mobius strip, the famous object used to attract fellow students to the wonders of topology. But the discoverer of this object, August Ferdinand Mobius was also responsible for the creation of another amazing thing, the theory of Mobius inversions. It is a formula in number theory that answers the following question: Given two functions $f$ and $g$ on the set of naturals, if it is known that $$f(n)=\sum_{d\vert n}g(d),$$ how do you express $g$ in terms of $f$? Actually, the Mobius inversion formula that is applied in number theory is a special case of a more general theorem in the branch of Order theory. Without any doubt, we explore the general theorem first! For that, we need to build up some theory though... POSETS: A poset is a short word for 'partially ordered sets'. It is a set $P$ equipped with a binary relation $\leq$ that satisfies the following three conditions: $\forall a\in P$, $a\leq a$ (Reflexivity

Preliminaries of the LTE Lemma

In this post I will discuss the preliminary theory underlying the Lifting the Exponent Lemma (let's refer it to as LTE from now on) which is  one of the foundational topics of Olympiad Number Theory.  It is a useful technique to think of whenever we come across Diophantine equations having exponents and it is even better if those exponents are primes. Comfort with modular arithmetic is assumed. If the reader is not fluent in Modular arithmetic, a quick reading from any handout is sufficient. (I highly recommend   this  ) So before starting things off, here are a few conventions that we will follow: We say an integer $b$ divides $a$ iff there is an integer $c$ such that $a = bc$. We denote it as $b \mid a$. We now take a look at more interesting things.  Consider the following: What is the greatest power of $3$ that divides $63$? It is easy to see that it's $2$. i.e $3^2 \mid 63$. We denote this as $\nu_3(63) = 2$. So let us now define it formally.  $\nu_p(x)$ is defined as the

Probability is Global

Today's blog post is an introduction to the Expected value and hopefully would be helpful to the kids who would be attending the upcoming Expected Value Lecture. The handouts/ books I referred to are Evan Chen's  Probability handout  , AOPS introduction to Counting and probability, Calt's  Expected value handout , brilliant and   this  IIT Delhi handout. Probability is Global Expected Value:    The  expected value  is the sum of the probability of each individual event multiplied by the number of times the event happens. It is denoted as $\Bbb E$ $[x]$ We have $$\Bbb E[x]=\sum x_n P(x_n)$$ where $x_n$ is the value of the outcome and $P(x_n)$ is the probability that $x_n$ occurs. Problem 1:  What is the expected value of the number that shows up when you roll a fair $6$ sided dice? Solution:  Since it's a fair dice, we get each outcome to have equal probability i.e $\frac {1}{6}.$ So $$\Bbb E[x]=\frac 16 \cdot 1+\frac 16 \cdot 2+\frac 16 \cdot 3+\frac 16 \cdot 4+\frac 16