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.

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## Comments

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TopNewestPlease Help.

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@Calvin Lin Can you help me with this question. It seems pretty difficult.

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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.Log in to reply

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Find \(\frac{1^2}{1!}+\frac{1^2+2^2}{2!}+\frac{1^2+2^2+3^2}{3!}+........\infty\)=?

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We can write the following as a summation

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

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do u have any generalisation for this \(\large{\displaystyle \sum^{\infty}_{r=0}\frac{r^n}{r!}}\)

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Is the answer \(\dfrac{17 \textbf{e}}{6}\)

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yes, plz post ur soln ???

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@Rajdeep Dhingra i'm waiting for ur solution

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@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

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

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

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

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

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

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

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

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

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OR use weighted AM-GM and then just AM-GM ._. Still like the unique approach. :o

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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.

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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

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The question is from GMO

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

Its wrong from the source itself.

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Comment deleted Jun 24, 2015

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Link

Wolfram Link. You can see there is no solution.

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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.

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There will be one such point also when XY will be passing through O

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prove that 11x31x61 divides 20^15-1

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Solve for x and y. x+xy+y=11 and xy(x+y)=30. Please provide a detailed solution.

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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.

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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 advanceLog in to reply

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.

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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,...)

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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].

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

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 ?

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Getting gold medal means 1st rank. Solve my set whose name is hard

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the required answer is positive , so what can be n, thats the question is asking for. IS there any possible answer.??

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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

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I have solved it !!!!!! I am Posting the solution.

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Good question. I show you where I have reached.

\[\text{We need to solve this equation to get the value of n} \\ \displaystyle \large n^2 = \omega^{2n-1}\]

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there is no +ve value of n

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