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RMO/INMO Doubt Board

Hi Guys since most of us are preparing for RMO everyone must be having doubts in some question or another. This board is open to all to ask/answer questions on topics included in RMO/INMO/IMO.

Do Reshare it so other members can see this Board.

Note by Rajdeep Dhingra
2 years, 2 months ago

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Is it possible to divide the first \(n\) prime numbers into two sets such that the sum of elements in each set is the same, if

a) \(n=2013^{2014}\)

b) \(n=2014^{2013}\) ?

Please Help. Samuel Jones · 2 years, 2 months ago

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@Samuel Jones @Calvin Lin Can you help me with this question. It seems pretty difficult. Samuel Jones · 2 years, 2 months ago

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@Samuel Jones It's not too hard, use techniques in Construction

Hint: 2 is the only even prime.

Hint: Bertrand's postulate tells us that between \( n \) and \( 2n \), there is a prime. Calvin Lin Staff · 2 years, 1 month ago

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Prove that for all positive integers \(n\) and \(k\), the greatest common divisor (g.c.d.) of the numbers \[ \displaystyle \binom{n}{k}, \binom{n+1}{k}, \binom{n+2}{k},...., \binom{n+k}{k}\] is \(1\).
Samuel Jones · 2 years, 2 months ago

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Find \(\frac{1^2}{1!}+\frac{1^2+2^2}{2!}+\frac{1^2+2^2+3^2}{3!}+........\infty\)=? Tanishq Varshney · 2 years, 1 month ago

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@Tanishq Varshney We can write the following as a summation

\[\sum_{k = 0}^{\infty}{\dfrac{n(n+1)(2n+1)}{6 n!}} \\\frac{1}{6}\sum_{k = 0}^{\infty}{\dfrac{2n^3 + 3n^2 + n}{n!}} \\ \frac{1}{3}\sum_{k=0}^{\infty}{\dfrac{n^3}{n!}} + \frac{1}{2}\sum_{k=0}^{\infty}{\dfrac{n^2}{n!}} + \frac{1}{6}\sum_{k=0}^{\infty}{\dfrac{n}{n!}} \\ e \left(\frac{5}{3} + \frac{2}{2} + \frac{1}{6} \right) \\ \boxed{\dfrac{17}{6}e}\] Rajdeep Dhingra · 2 years, 1 month ago

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@Rajdeep Dhingra do u have any generalisation for this \(\large{\displaystyle \sum^{\infty}_{r=0}\frac{r^n}{r!}}\) Tanishq Varshney · 2 years, 1 month ago

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@Tanishq Varshney Is the answer \(\dfrac{17 \textbf{e}}{6}\) Rajdeep Dhingra · 2 years, 1 month ago

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@Rajdeep Dhingra yes, plz post ur soln ??? Tanishq Varshney · 2 years, 1 month ago

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@Rajdeep Dhingra @Rajdeep Dhingra i'm waiting for ur solution Tanishq Varshney · 2 years, 1 month ago

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@Tanishq Varshney @Tanishq Varshney I will write just write the function then u can find its Summation.

Summation of first n squares = n(n+1)(2n+1)/6 so the above series can be written as

n(n+1)(2n+1)/6(n!). Find its summation answer will be 17e/6 Kalash Verma · 2 years, 1 month ago

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@Kalash Verma u mean \(\frac{n(n+1)(2n+1)}{6n!}\) Tanishq Varshney · 2 years, 1 month ago

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@Tanishq Varshney Yeah, sorry for typo edited. Kalash Verma · 2 years, 1 month ago

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If a,b,c are positive real numbers such that abc=1 Prove that \(a^{b+c}b^{a+c}c^{a+b}\leq1\)
Shivam Jadhav · 2 years, 2 months ago

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@Shivam Jadhav First, consider the following Lemma,

Lemma: For positive \(x\) and \(y\), the following inequality holds,

\[xy(x\ln x + y\ln y) \geq \ln x + \ln y\]

The equality case being \(x=y=1\)

Proof: We'll divide the proof into three cases,

Case 1: \(x,y \geq 1\)

\(\implies \ln x,\ln y \geq 0\)

\(\implies x\ln x \geq \ln x\)

\(\text{&}\)

\(y\ln y \geq \ln y\)

\(\implies x\ln x + y\ln y \geq \ln x +\ln y\)

\(\implies xy(x\ln x +y\ln y) \geq \ln x + \ln y\) \(( \because x,y \geq 1)\)

Similarly, we can proceed with other two cases, i.e.,

Case 2: \(x,y < 1\)

Case 3:\(x<1, y>1\)

This proves our Lemma. \(\square\)

Now, using the Lemma, we have,

\(ab(a\ln a + b\ln b) \geq \ln a + \ln b\)

\(\implies \ln a \times \left(a-\dfrac{1}{ab}\right) + \ln b \times \left(b-\dfrac{1}{ab}\right) \geq 0\)

\(\implies (a-c) \times \ln a + (b-c) \times \ln b \geq 0\) \((\because abc=1)\)

\(\implies a\ln a + b\ln b -c \times (\ln a +\ln b) \geq 0\)

\(\implies a\ln a +b\ln b +c\ln c \geq 0\) \((\because abc=1 \implies \ln a + \ln b + \ln c = 0)\)

\(\implies \ln (a^a b^b c^c) \geq 0\)

\(\implies a^a b^b c^c \geq 1\)

\(\implies a^{b+c} b^{a+c} c^{b+a} \leq 1\)

Q.E.D. Ishan Singh · 2 years, 2 months ago

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@Ishan Singh OR use weighted AM-GM and then just AM-GM ._. Still like the unique approach. :o Kunal Verma · 1 year, 5 months ago

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@Shivam Jadhav Here's a much simpler solution -- We can rewrite the LHS of the inequality as (ac)^b (ab)^c (bc)^a = 1/(a^a b^b c^c) (replacing all the ab's with 1/c's etc.). Now suppose, WLOG, a >= b >= c. Since they multiply to one, a >=1 and c <=1, hence a^a >= a^b (because b <= a), c^c >= c^b, hence a^b b^b c^c >= a^b b^b c^b = 1. Thus 1/(a^a b^b c^c) <= 1. Abhimanyu Pallavi Sudhir · 5 months, 2 weeks ago

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If x and y are positive real numbers . Prove that \(4x^{4} + 4y^{3} + 5x^{2} +y + 1 \geq 12xy\)
Shivam Jadhav · 2 years, 2 months ago

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@Shivam Jadhav Clearly, for positive \(x\) and \(y\),

\(4x^{4} + 4y^{3} + 5x^{2} +y + 1 > 6x^2 +6y^2 \geq 12xy\) (By A.M. - G.M. inequality)

I don't think the equality holds though. Kindly check the question again @Shivam Jadhav Ishan Singh · 2 years, 2 months ago

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@Ishan Singh The question is from GMO Shivam Jadhav · 2 years, 2 months ago

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@Ishan Singh It doesn't. That's why I got this question wrong. >.<

Its wrong from the source itself. Siddhartha Srivastava · 2 years, 2 months ago

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Comment deleted Jun 24, 2015

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@Rajdeep Dhingra Link

Wolfram Link. You can see there is no solution. Siddhartha Srivastava · 2 years, 2 months ago

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Solve for x and y. x+xy+y=11 and xy(x+y)=30. Please provide a detailed solution. Arkodipto Dutta · 9 months, 3 weeks ago

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Question:Consider two circles S and R.Let the center of circle S lie on R.Let S and R intersect at A and B.Let C be a point on S such that AB=AC.Then prove that the point of intersection of AC and R lies in or on S. Hemant Kumae · 1 year, 3 months ago

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hi guys if you mind my question, how to become like you guys? i mean really good at mathematics? I ranked no.1 in our town and school (btw, my real age is 16) but i dont know where to start. I really want to solve those olympiad-like problems out there... Thanks in advance すすべての すべての · 1 year, 7 months ago

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@すすべての すべての Try out books like Mathematical Olympiad Challenges by Titu Andreescu or Mathematical Olympiad treasures by the same, Problem solving by Arthur Engel, Rajeev Manocha, Thrills of Pre-College Mathematics( haven't really used it but people say it's good ). Keep practicing RMO papers over and over again and if you have the courage, try out questions from China's Olympiads too. Kunal Verma · 1 year, 5 months ago

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Find all natural n > 1 for which sum \(2^2 + 3^2 + .... + n^2\) equals to \(p^k\) where p is prime and k is natural.

also find gcd(2002 + 2, 2002^2 + 2, 2003^3 + 2,...) Dev Sharma · 1 year, 8 months ago

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[Z] is complex number. [ |Z| = |Z+1| = 1 ] and [ Z^ {2} = (Z +1) n] Here [n] is a positive integer. find minimum value of [n]. Shivam Jadhav · 2 years, 1 month ago

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ω is cube root of unity What is the smallest positive integer that satisfies the given equation \[ω^{ n }=(1+ω)n\]
Shivam Jadhav · 2 years, 2 months ago

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@Shivam Jadhav \[(1+ \omega)n = \omega^n \\ \text{squaring both sides} \\ (1 + \omega^2 + 2\omega) n^2 = \omega^{2n} \\ \omega n^2 = \omega^{2n} \\ \text{cubing both sides } \\ \omega^3 n^6 = \omega^{6n} \\ \rightarrow n^6 = 1 \\ \text{The smallest solution of this over the reals and satisfying the equation is -1 , so} \\\displaystyle n = \boxed{-1} \]


Just curious : How did you get IMO rank 1. I mean do they tell the Rank. I thought they just gave gold medal.
BTW Congratulations for making it to INMO. Any guidance you could give me ? Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra Getting gold medal means 1st rank. Solve my set whose name is hard Shivam Jadhav · 2 years, 2 months ago

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@Shivam Jadhav Any guidance for me ? Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra Use book excursion in mathematics Shivam Jadhav · 2 years, 2 months ago

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@Shivam Jadhav cant find it on flipkart Dev Sharma · 1 year, 8 months ago

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@Shivam Jadhav Anything other ? Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra the required answer is positive , so what can be n, thats the question is asking for. IS there any possible answer.?? Tanishq Varshney · 2 years, 2 months ago

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@Tanishq Varshney Well there is no possible positive answers as I think the solutions to my equation is 1 or -1. Trying 1 is not giving the correct answer so -1 is what we are left with. Rest are complex solutions. Rajdeep Dhingra · 2 years, 2 months ago

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@Shivam Jadhav is there any answer to this?? its very well clear that -1 satisfies the equation but how to find the smallest possible integer, i am clueless Tanishq Varshney · 2 years, 2 months ago

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@Tanishq Varshney I have solved it !!!!!! I am Posting the solution.

Rajdeep Dhingra · 2 years, 2 months ago

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@Shivam Jadhav Good question. I show you where I have reached.
\[\text{We need to solve this equation to get the value of n} \\ \displaystyle \large n^2 = \omega^{2n-1}\] Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra there is no +ve value of n Karan Siwach · 2 years, 2 months ago

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@Karan Siwach That's what even I said in the above comment in which I have posted the solution. Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra i didn't read the comments carefully, sorry for that :) Karan Siwach · 2 years, 2 months ago

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@Karan Siwach No Problem. Rajdeep Dhingra · 2 years, 2 months ago

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AB and CD are 2 straight lines meeting at O and XY is another straight line. Show that in general 2 points can be found in XY which are equidistant from AB and CD . When is there only one such point ?
Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra Draw the angle bisectors of the lines AB and CD. Let these lines be \(L_{1}\) and \(L_{2}\).

Then in general, XY will cut each of these lines in one point each. These two points are equidistant from AB and CD as they lie on the angle bisectors.

There will be only one point if XY is parallel to either of \(L_{1}\) or \(L_{2}\) and thus intersects only one of them. Shashwat Shukla · 2 years, 2 months ago

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