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 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 😎). *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 😊.

---

I also made a pdf compilation of these problems and hints. Here is the google drive link https://drive.google.com/file/d/1sFQBU1MDIYGXWKG_1Bb6TpUAfTZfT4en/view?usp=sharing

Sunaina 💜