Problems with meeting people!

Yeah, I did some problems and here are a few of them! I hope you guys try them!

Putnam, 2018 B3

Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n - 1$, and $n-2$ divides $2^n - 2$.

Proof

We have $$n|2^n\implies n=2^a\implies 2^a-1|2^n-1\implies a|n\implies a=2^b$$ $$\implies 2^{2^b}-2|2^{2^a}-2\implies 2^b-1|2^a-1\implies b|a\implies b=2^c.$$ Then simply bounding.

USAMO 1987

Determine all solutions in non-zero integers $a$ and $b$ of the equation $$(a^2+b)(a+b^2) = (a-b)^3.$$

Proof

We get  $$2b^2+(a^2-3a)b+(a+3a^2)=0\implies b = \frac{3a-a^2\pm\sqrt{a^4-6a^3-15a^2-8a}}{4}$$ $$\implies a^4-6a^3-15a^2-8a=a(a-8)(a+1)^2\text{ a perfect square}$$ $$\implies a(a-8)=k^2\implies a^2-8a-k^2=0\implies \implies a=\frac{8\pm\sqrt{64+4k^2}}{2}=4\pm\sqrt{16+k^2}.$$ $$16+k^2=m^2\implies (m-k)(m+k)=16.$$

Now just bash.

USAMO 1988

Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum $$|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|$$ ends in the digit $9$.

Proof

Note we just have to show that the answer is $-1 \mod 10.$ We will show that the answer is $1\mod 2,~~4\mod 5.$ Note that $a_i-b_i$ and $a_i+b_i$ have the same parity. \\ So the sum is $$\sum (a_1+b_i)\mod 2 \equiv \frac{1998\cdot 1999}{2}\mod 2.$$ And $|a_i-b_1|$ is always $1\pmod 5.$ Hence the sum is $999\cdot 1 \equiv 4\mod 5.$

IMO 2009 P1

Let $n$ be a positive integer and let $a_1,a_2,a_3,\ldots,a_k$ $( k\ge 2)$ be distinct integers in the set ${ 1,2,\ldots,n}$ such that $n$ divides $a_i(a_{i + 1} - 1)$ for $i = 1,2,\ldots,k - 1$. Prove that $n$ does not divide $a_k(a_1 - 1).$

Proof

Assume not, then $a_ka_1 \equiv a_k \mod n.$ The given conditions can be written as, $a_i\cdot a_{i+1}\equiv a_i\mod n.$ So $$a_1\cdot a_2\equiv a_1\mod n$$ $$a_2\cdot a_3\equiv a_2\mod n$$ $$\vdots$$ $$a_{k-1}\cdot a_k\equiv a_{k-1}\mod n$$ So we have $$a_1\equiv a_1\cdot a_2\equiv a_1\cdot a_2\cdot a_3\equiv \dots \equiv a_1\cdot\dots\cdot a_k\mod n.$$ Now, we again reduce, so we get $$a_1\cdot (a_2\dots\cdot a_{k-2}\cdot a_{k-1})\cdot a_k\equiv a_1\cdot (a_2\dots\cdot a_{k - 2})\cdot a_k\equiv\dots\equiv a_1\cdot a_{k}\equiv a_k\mod n$$ So we get $a_1\equiv a_k\mod n.$ Which is not possible, since $a_1,a_k\in{1,\dots n}$ and are distinct.

Iberoamerican 2020 P2

Let $T_n$ denotes the least natural such that $$n\mid 1+2+3+\cdots +T_n=\sum_{i=1}^{T_n} i$$ Find all naturals $m$ such that $m\ge T_m$.

Proof

Note that all odd $m$ works. For $n=2^k$, it is not possible. Let $n=2^kp_1^{\alpha_1}\cdots p_l^{\alpha_l}$. We want $$M(M+1)\equiv 0\mod 2n.$$ And we want $M\le m.$ Note that such number satisfies $M(M+1)\equiv 0\mod 2^{k+1}\prod p_i^{\alpha_i}$ then $$M\equiv 0 \text { or } 1 \mod 2^{k+1}, p_i^{\alpha_i}.$$ By CRT, we get that there are $2^{l+1}$ such numbers. Now, if any numbers are less than $m$ then we are done. Also, note that these numbers range from $0$ to $2m$. Note that if $$r(r+1)\equiv 0\mod 2m\implies (2m-1-r)(2m-1-r+1)\equiv 0\mod 2m.$$ So if any of these number $M>m,$ simply take $2m-1-M< m$ (and $M\ne 2m-1$)

IMO 2005 P6

Determine all pairs $(x, y)$ of integers such that $$1+2^{x}+2^{2x+1}= y^{2}.$$

Proof

Note that $y$ is odd. Clearly $(x,y)=(0,\pm 2)$ satisfies. Assume $x\ge 1$. Then by $\mod 8$, we get $x\ge 3.$ Note that $$2^x(2^{x+1}+1y^2-1=(y-1)(y+1).$$ Note that $$\gcd(y-1,y+1)=2.$$ We have two cases. 

Case 1: $v_2(y-1)=1,v_2(y+1)=x-1$ 

Proof: Then let $y=2^{x-1}\cdot k+1,~~k\text{ is odd}$ and $2^{x-1}=a$. We get $$2a(4a+1)=(ka+1)^2-1 \implies (8-k^2)a=2k-2.$$ We get that there is no such $k$. 

Case 2: $v_2(y+1)=1,v_2(y-1)=x-1$

Proof: So $y=k(2^{x-1})-1$ and $k$ is odd. Let $a=2^{x-1}$. We get $$2a(4a+1)=(ka-1)^2-1\implies 2+2k=(k^2-8)a.$$ Note that $k^2-8>2+2k$ for $k\ge 5$ and clearly $k>2$. So $k=3$ works. For $k=3$ we get $(x,y)=(4,\pm 23).$ So the answer is $\boxed{ (0,\pm 2),(4,\pm 23)}.$

Jappan 1996

Let $m,n$ be positive integers with $(m,n)=1$. Find $(5^m+7^m,5^n+7^n)$, given $m,n$ odd.

Proof

Let $\gcd(5^m+7^m,5^n+7^n)=d$. Now note that $$5^m\equiv -7^m\mod d$$ $$5^n\equiv -7^n\mod d.$$ So $$\text{ord}_{d}(5/7)|2m, ~~\text{ord}_{d}(5/7)|2n\implies \text{ord}_{d}(5/7)=1\text{ or } 2.$$ So $$5^2\equiv 7^2\mod d,~~d|7^2-5^2=24.$$ But $$v_2(5^m+2^m)=2,~~v_3(5^m+7^m)\ge 1\implies \gcd(5^m+7^m,5^n+7^n)=12.$$

2006 N2

For $x \in (0, 1)$ let $y \in (0, 1)$ be the number whose $n$-th digit after the decimal point is the $2^{n}$-th digit after the decimal point of $x$. Show that if $x$ is rational then so is $y$.

Proof

Given $x$ is rational. So $x$ would be eventually periodic. Let the periodic length be $l$. \newline Now, note that if $\gcd(l,2)=1$, then $y$ will be eventually periodic $\mod \phi(l)$ as $\exists N$ such that we have $$2^n\equiv 2^{n+\phi(l)}\mod l\text{ for all } n>N$$ If $\gcd(l,2)=2$, then $y$ will be eventually periodic $\mod \phi(\frac{l}{v_2(l)})$ as $\exists N$ such that we have $$2^n\equiv 2^{n+\phi(\frac{l}{v_2(l)})}\mod l\text{ for all } n>N.$$

2017 EGMO P6

$ABC$ in its sides $BC,CA,AB$ are denoted by $G_1,G_2,G_3$ and $O_1,O_2,O_3$, respectively. Show that the circumcircles of triangles $G_1G_2C$, $G_1G_3B$, $G_2G_3A$, $O_1O_2C$, $O_1O_3B$, $O_2O_3A$ and $ABC$ have a common point.

Proof

Define $X$ as the anti-Steiner point of Euler line. Moreover, introduce $H$, $H_1$, $H_3$ analogously. We claim that all these circles pass through $X$. Note that $BXH_1H_3$ cyclic. So we have $$\angle H_1XH_3=\angle H_1BH_3=\angle H_1BH+\angle HBH_3=2(90-C)+2(90-A)=360-2(A+C)=2B.$$ As $H_1-G_1-O_1,H_3-G_3-O_3$ collinear, we have $$\angle O_1XO_3=2B=\angle G_1XG_3.$$ But note that, due to the reflection, we have $$\angle O_1BO_3=2B=\angle G_1BG_2\implies O_1XO_3B, G_1XG_3B \text{ cyclic}.$$

That's it for the math part. Here comes the non-math part!

Woah it's been months since I posted on this blog!!!!! Darn, there's a lotttt to update about my life! 

I (and Atul) were accepted for the Maths Beyond Limits camp which was going to be held in Poland. So for the Visa stuffs, I met Swastika and Atul!! 

Kolkata

When I was in Kolkata, I went to ISI Kolkata! Woah it is such a beautiful campus! And I met Neena Gupta there and talked for a good 45 mins and she explained to me how important it is to continuously work no matter where I am studying. She was so down to earth and is so good at conversations ( I was nervous the whole time and I was screaming inside AAAAAAAAAAAAAAAAAAAAAAAA)

Kolkata and Guwahati, weather-wise don't have much difference. As in, both cities have almost same humidity etc. Even the people there, they talk in bangla so yeah I can survive in Kolkata because I understand Bangla and the food style is also similar. 

The buildings are huge though, like the area near the Kolkata airport is so so so well developed, very beautiful! 

Since I was in Kolkata, I and Swastika decided to meet up in the Eco park. Her house is quite close to Eco park and Kolkata airport is close too ( We had flight the same night). 

And WOW. ECO PARK IS HUGE. And it's so pretty! I and Swastika were lost in park, it's so huge! We talked a lot and her brother is insanely smart and cute!

A few photo dumps! 

Bangalore

Bangalore was hectic. We had planned to stay two nights one day in Bangalore. Anyways, Atul and I decided to meet and come to my place. 

HE IS SO TALL!!!!!!!! ( LIKE HE IS) 

He played chess with Suhan and during the match, I and my cousin were so bored.  But then we decided to go to Dominoes and yeah that was fun!!! I still haven't accepted the fact that someone can like onions on pizzas ( onions in general are just nasty).. We talked about a lot of stuffs ( all non math) and yeah the pizzas were great! Our talks were mainly me talking, him listening and then interrupting me and ask me to eat my pizza. (Yes I talk a lot when I am with the right person)

We then decided to just walk back from the dominos place to my home and then I realised dogs are scary and Atul is scarier. THERE WAS A DOG IN OUR WAY. WHAT'S THE BEST THING TO DO? CHANGE OUR LANE right? WHAT THIS GUY DID WAS DRAG ME TO THE DOG'S WAY..

Then I realised Atul's plan was not to meet me but was to take revenge because of how much I annoyed him. Oki oki Just kidding, Atul is a very nice person and the bestest friend I ever had. 

We, unfortunately, didn't take any selfies ( Your fault Atul, you toh are not careless nah, I know I am very careless, then why did you not remind me for taking selfie) but we had one pic. Let me add that and then end this section! 

We both are not looking the best but it's fine

Technotholon

I and Ananda met Rushil, Krutarth and Debayu!! So a few photo dumps!

And we even watched stand-up comedy where we had to pretend we were long-lost siblings!! 

EGMOTC

It was a fun experience meeting new people! Writing EGMOTSTs were stressful too! We had late night dance parties, gossips about so many random stuffs! 

I and Gunjan were room mates! ( Gunjan's reaction was memorable cause the whole pre camp, she was like "I want a single room, Sunaina", guess it's faith?) We actually met in the airport and darn she is so pretty!!! Perfect example of beauty with brain! 

We had to find our cab and it was so hard and plus Gunjan was talking to the cab driver and she heard " Green Indian Flag".. ( For context, near the Chennai Airport is huge Indian flag where all Taxi drivers stand)

On our way going to CMI, we were talking with few of our friends ( Malay, Pranav and Atul). When we reached CMI, I met Rohan Bhaiya! And then Aditi di! 

My first reaction to CMI was "Ooo warm colours and greenery".. CMI has blackboards every where ( students can discuss about problems legit anywhere).

We then met Yashwi and Bhavya, who became roomates! Sanjana came the next day ( and OMG she is so tiny). Sanjana got the single room! Gunjan wanted single room but I emotionally convinced her to stay in our room.. 

We met Ananya di, Rohinee di, Prateek bhaiya and Sahil bhaiya the next day!

Apart from our first night where we simply slept, rest all the nights we( I and Gunjan) talked so so so much! Gunjan has such a fun personality and we had a lot of common friends so both of us could understand each other!

From Dec 2, our classes started and omg they were so tiring, I got so busy that I wasn't able to text people back.. We had classes from 8:30 to 6 ( with lunch break) and then zeta time from 10pm to whatever time we wanted. 

In one zeta time, Rohan bhaiya and Sahil bhaiya somehow convinced most of us that Reals are countable and I was like " NO REALS ARE NOT COUNTABLE" but then even I started questioning my life. 

I missed out on the second zeta time but from my friends reaction, I could see they were questioning their life.. It was on " Axiom of choice". 

The classes were interactive, not heavy on theory and a lot of problems. Rohan bhaiya had a lot to talk on Ramsey theory, which for sure blown our mind! "Edges in multidimensional and proving theorems on those dimensions".. Fancy RG ( but yes the ideas were nice)

All the classes and professor interactions were so fun! 

We had a dance party where even Ananya di joined us! ( And we danced in a lot of Bollywood songs)

On Dec 4, we went to Marina Mall, and brought a lot of stuffs and chocolates for Bhavya ( cause it was her birthday!) and then we had a practice test! I got very lucky in my practice test and scored 7|7|1. Problems were graded out of 7 here btw. Rg said I could have arbitrated and could have got more marks for P3 but it's fine.. I was happy only..

Dec 5-9 we were having classes and then I was going to the library to study ( BTW CMI Library is great)..

We girls were having so much fun!! 

Then we had our TSTs happen and I got my score as 3|0|3 for day 1 and 1|6|10 for day 2  TSTs were graded out of 10. 

P6 was proposed by Atul!!! On Day 2, Once I submitted my solutions to Rohan bhaiya and I was happy that I fully solved P6, Rohan Bhaiya said "guess who proposed P6?" and I was like "who?" and he said "Your best friend". And I was surprised as hell. 

Because I had told him to not propose for this year's TSTs as it's embarrassing to not solve your own best friend's problem. ( Atul you betrayed me )

Anyways, congrats Atul for the problem, it's a nice problem!

We played pictionary and codenames. Oh yes, we also had Estimathon and "solve it in your mind". These were so fun!  We talked about K-Pop too in a lecture! 

We were just too lazy to click pictures actually

Thanks to Rohan bhaiya for treating us to such delicious food. The EGMO TC was just 11 days camp but it made all of us so close.

Thanks to EGMOTC girls for taking care of me when I was sick!

Thanks to Rohan bhaiya, CMI, PD sir, the proposers and all the instructors for making EGMO TC possible! 

Thanks to Malay, Pranav, Sid, and Swastika for wishing me luck and being nice friends. Thankyou to Krutarth, Ananda and Rushil too ( even though they forgot to wish me luck) for gsolving with me so much!

And obviously, Atul, thank you for listening to me even though you were busy with College stuff, having so many  conversations, calling me whenever I needed you,  being such a nice friend, motivating me to have confidence and thanks for believing in me . Although you weren't physically there in the camp with me, you still played such an important part in the camp for me..( Oki too much emotional stuffs, but yes you are the best, dum kid. )

oh btw, yes, I did make the EGMO team.