### Top 10 problems of the week!

This week was full Number Theory and algebra biased!! ðŸ˜„

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 College math but it's great! )

a. FTSOC let $a,b,c$ be primes. Then note that by simon's favorite factoring trick , we get $(b+1)(c+1)\equiv 1 \mod a$ , similarly for $b,c$.

b. Bound $\frac{(a+1)(b+1)(c+1)}{abc}$ and get a contradiction !

3rd position (IMO 2011 P1): Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \leq i < j \leq 4$ for which $a_i +a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$.

Walkthrough: This was clearly the hardest of all like I took  lot of hints :P.

a. Show that $n_A\ne 6,5$.

b. We will show that $n_A=4$ is possible. hmm...so what do we have ?

c. Show that $a_1+a_4=a_2+a_3$.

d. So $a_1+a_2|2(a_2+a_3) \implies k_1(a_1+a_2)=2(a_2+a_3)$ , and similarly $k_2(a_1+a_3)=2(a_2+a_3)$ , where $k_1>k_2$.

e.But note that $2a_3+2a_1>2(a_2-a_1) \implies k_2=3$.

f. Show that $k(a_1+a_2)=6a_2-6a_1 \implies k_1=4,5$

g. Case bash!!!!!!!!!

2nd position(IMO 2011 P5) : Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m- n)$. Prove that, for all integers $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m)$.

Walkthrough: Thanks to the Pr0est Mueller.25

a. Take $P(m,0)$ and show that $f(m)|f(0)$ for all $m$. This gives that $f(x)$ has finite solutions.

b. Take $P(0,n)$ and $P(0,-n)$ and show that $f(n)=f(-n)$.

c. Prove by induction that if $a|b \implies f(a)|f(b)$ [it's not required, we just want $f(1)|f(b)$, but it's cute, so one can try].

d. Since $f(x)$ has finite solution, let the solutions be $f(1)<f(a_1)< \dots <f(a_k)<f(0)$.

e. So.. the problem now reduces on showing $f(a_i)|f(a_j)$ when $i<j$.

f. okay so we have $f(1)|f(a_i)$ , so let's try showing $f(a_1)|f(a_2)$.

g. Let's take $P(a_2,a_1)$ then show that $f(a_2-a_1)= f(a_1)$ or $f(1)$.

h. But we want to show that $f(1)|f(2)$ . But note that if we show that $f(a_2-a_1)= f(a_1)$, then we will be done! So let's try to show that!

i. Take $P(a_2-a_1, -a_1)$ and wola!

1st Position (IMO Shortlist 2011 N3):  Let $n \geq 1$ be an odd integer. Determine all functions $f$ from the set of integers to itself, such that for all integers $x$ and $y$ the difference $f(x)-f(y)$ divides $x^n-y^n.$

Walkthrough: a. Try to guess what possible solutions can be by taking $n=1,3,6$ ( just guess, no need to prove :P)

b. Note that if $f(x)$ works then $f(x)+c, -f(x)$ works too. So we can assume $f(0)=0$ and $f(1)=1$.

c. With $P(p,1)$ and $P(p,0)$, where $p$ is a prime, show that $f(p)=\pm p^k$ , where $k|n$

d. Show $f(p)\ne -p^k$ [ we here use the fact that when $a^b-1|a^c-1 \implies b|c$ and also $k\ne 0$ ]

e. But n has finitely many prime factors , and there are finitely many prime, so there will exist a prime factor of $n$ say $d$ , which will be used infinitely times. So let $q$ be a very( verrrry big) prime with $f(q)=q^d$

f. Take $P(x,q)$, and show that $f(x)=x^d$. And then conclude!

g. Solutions are $\boxed{f(x)=\pm x^d+c}$ where $d|n$.

Next are the walkthroughs of this week's top 5 Algebra problems! This is only for beginners algebra people ( It's not my fault that people who are reading this blog are pr0s ðŸ˜Ž). *Take it more like as a set of problems to motivate you to study algebra :P

5th position (2010 AMC 10 A P21) : The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$?

Walkthrough: a. Use Vieta's formula and factorize $2010=2\cdot 3 \cdot 5\cdot 67$

b. okay to minimize , obviously 6,5,67 will be answer :P (idk how to explain this part , it actually follows trivially :P)

4th position (NIMO summer contest P9): The roots of the polynomial $P(x) = x^3 + 5x + 4$ are $r, s$, and $t$. Evaluate$(r + s) ^4 (s + t) ^4 (t + r)^ 4$

Walkthrough: a. Use Viteta's , we get $r+s+t=0$

b. we just have to find $(rst)^4$.

3rd position (AIME 2008 II P7): Let $r$, $s$, and $t$ be the three roots of the equation $8x^3+1001x+2008=0.$ Find $(r+s)^3+(s+t)^3+(t+r)^3$.

Walkthrough: Ooo quite similar to the problem we did  previously.

a. Use viteta and get $r+s+t=0$. So $(r+s)^3+(s+t)^3+(t+r)^3=-(t^3+r^3+s^3)$.

b. So for people who are in grade 9 or below India standard, or any beginner in algebra, there's a very well known formula, which says $a^3+b^3+c^3=3abc$ , when $a+b+c=0$, it's just $a^3 + b3^ + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ . Use it and we are done!

2nd position (2017 AMC 12 A P17) : There are $24$ different complex numbers $z$ such that $z^{24}=1$. For how many of these is $z^6$ a real number?

Walkthrough:  I don't think so any walkthrough is needed :P. Answer is $\boxed{12}$. Consider $z^{{\frac{2\pi\cdot i \cdot k}{24}}\cdot 6}$ , it will be real iff $k$ is even .

1st Position(JBMO 2012 Shortlist): Let $a$ , $b$ , $c$ be positive real numbers such that $abc=1$ . Show that :

$\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab} \leq \frac{ \left (ab+bc+ca \right )^2 }{6}$

Walkthrough: a. Use AM-GM and get $a^3+bc\ge 2a$.

b. Try to get this $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{(ab+bc+ca)^2}{3}$ , conclude with AM-GM again !

So these were my top 10 ! I personally wanted to add JMO 2017 P4, but didn't ( due to some soul imbalance , you will understand what I mean once you try this problem) . Go ahead and try if u want to :P.

What are your top 10s ? do write in the comments section (at least write something ! I will be happy to hear your comments ). Follow this blog if you want to see more contest math problems! See you all soon ðŸ˜Š.

---

Sunaina ðŸ’œ

### The importance of "intuition" in geometry

Hii everyone! Today I will be discussing a few geometry problems in which once you "guess" or "claim" the important things, then the problem can easily be finished using not-so-fancy techniques (e.g. angle chasing, power-of-point etc. Sometimes you would want to use inversion or projective geometry but once you have figured out that some particular synthetic property should hold, the finish shouldn't be that non trivial) This post stresses more about intuition rather than being rigorous. When I did these problems myself, I used freehand diagrams (not geogebra or ruler/compass) because I feel that gives a lot more freedom to you. By freedom, I mean, the power to guess. To elaborate on this - Suppose you drew a perfect  diagram on paper using ruler and compass, then you would be too rigid on what is true in the diagram which you drew. But sometimes that might just be a coincidence. e.g. Let's say a question says $D$ is a random point on segment $BC$, so maybe

### 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 what? I was not alone, Krutarth too fakesol

### Edge querying in graph theory

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