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).
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)
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 😊.
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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 💜
Nice Post!
ReplyDeleteBTW which year's APMO was N4? Asking 'cause its almost same as IMO 1992/P1
I actually don't know.. Evan sourced it as APMO in his orders modulo prime handout ..
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