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.

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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TopNewestPlease Help. – Samuel Jones · 2 years, 2 months ago

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

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Construction

It's not too hard, use techniques inHint: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 agoLog in to reply

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

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

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– Tanishq Varshney · 2 years, 1 month ago

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

– Rajdeep Dhingra · 2 years, 1 month ago

Is the answer \(\dfrac{17 \textbf{e}}{6}\)Log in to reply

– Tanishq Varshney · 2 years, 1 month ago

yes, plz post ur soln ???Log in to reply

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

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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 – Kalash Verma · 2 years, 1 month ago

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– Tanishq Varshney · 2 years, 1 month ago

u mean \(\frac{n(n+1)(2n+1)}{6n!}\)Log in to reply

– Kalash Verma · 2 years, 1 month ago

Yeah, sorry for typo edited.Log in to reply

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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 agoLog in to reply

– Kunal Verma · 1 year, 5 months ago

OR use weighted AM-GM and then just AM-GM ._. Still like the unique approach. :oLog in to reply

– Abhimanyu Pallavi Sudhir · 5 months, 2 weeks ago

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.Log in to reply

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\(4x^{4} + 4y^{3} + 5x^{2} +y + 1 > 6x^2 +6y^2 \geq 12xy\) (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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– Shivam Jadhav · 2 years, 2 months ago

The question is from GMOLog in to reply

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

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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 agoLog in to reply

– Kunal Verma · 1 year, 5 months ago

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.Log in to reply

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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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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– Shivam Jadhav · 2 years, 2 months ago

Getting gold medal means 1st rank. Solve my set whose name is hardLog in to reply

– Rajdeep Dhingra · 2 years, 2 months ago

Any guidance for me ?Log in to reply

– Shivam Jadhav · 2 years, 2 months ago

Use book excursion in mathematicsLog in to reply

– Dev Sharma · 1 year, 8 months ago

cant find it on flipkartLog in to reply

– Rajdeep Dhingra · 2 years, 2 months ago

Anything other ?Log in to reply

– Tanishq Varshney · 2 years, 2 months ago

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

– Rajdeep Dhingra · 2 years, 2 months ago

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.Log in to reply

– Tanishq Varshney · 2 years, 2 months ago

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 cluelessLog in to reply

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\[\text{We need to solve this equation to get the value of n} \\ \displaystyle \large n^2 = \omega^{2n-1}\] – Rajdeep Dhingra · 2 years, 2 months ago

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– Karan Siwach · 2 years, 2 months ago

there is no +ve value of nLog in to reply

– Rajdeep Dhingra · 2 years, 2 months ago

That's what even I said in the above comment in which I have posted the solution.Log in to reply

– Karan Siwach · 2 years, 2 months ago

i didn't read the comments carefully, sorry for that :)Log in to reply

– Rajdeep Dhingra · 2 years, 2 months ago

No Problem.Log in to reply

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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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