# 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
4 years, 7 months 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 $1$.

- 4 years, 7 months ago

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) $n=2014^{2013}$ ?

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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.

Staff - 4 years, 7 months ago

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 ?

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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

- 1 year, 4 months ago

If x and y are positive real numbers . Prove that $4x^{4} + 4y^{3} + 5x^{2} +y + 1 \geq 12xy$

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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

Its wrong from the source itself.

- 4 years, 7 months ago

The question is from GMO

- 4 years, 7 months ago

If a,b,c are positive real numbers such that abc=1 Prove that $a^{b+c}b^{a+c}c^{a+b}\leq1$

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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

- 3 years, 11 months 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.

- 2 years, 11 months ago

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

- 4 years, 7 months ago

We can write the following as a summation

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

- 4 years, 7 months ago

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

- 4 years, 7 months ago

Is the answer $\dfrac{17 \textbf{e}}{6}$

- 4 years, 7 months ago

yes, plz post ur soln ???

- 4 years, 7 months ago

@Rajdeep Dhingra i'm waiting for ur solution

- 4 years, 7 months ago

@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

- 4 years, 7 months ago

u mean $\frac{n(n+1)(2n+1)}{6n!}$

- 4 years, 7 months ago

Yeah, sorry for typo edited.

- 4 years, 7 months ago

ω is cube root of unity What is the smallest positive integer that satisfies the given equation $ω^{ n }=(1+ω)n$

- 4 years, 7 months ago

Good question. I show you where I have reached.
$\text{We need to solve this equation to get the value of n} \\ \displaystyle \large n^2 = \omega^{2n-1}$

- 4 years, 7 months ago

there is no +ve value of n

- 4 years, 7 months ago

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

- 4 years, 7 months ago

- 4 years, 7 months ago

No Problem.

- 4 years, 7 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 clueless

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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

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 ?

- 4 years, 7 months ago

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

- 4 years, 7 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.

- 4 years, 7 months ago

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

- 4 years, 7 months ago

Any guidance for me ?

- 4 years, 7 months ago

Use book excursion in mathematics

- 4 years, 7 months ago

Anything other ?

- 4 years, 7 months ago

cant find it on flipkart

- 4 years, 2 months ago

[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].

- 4 years, 7 months ago

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,...)

- 4 years, 2 months ago

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

- 4 years, 1 month 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.

- 3 years, 11 months ago

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.

- 3 years, 9 months ago

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

- 3 years, 3 months ago

prove that 11x31x61 divides 20^15-1

- 2 years, 1 month ago