×

# RMO board

Hi guys!

I know that many of you must be RMO aspirants, and are preparing tough for that. But even all of us know that RMO is not that easy to qualify.There are a lot of problems to do and concepts to learn.So why not discuss and gain more and more knowledge?

This board has been made for that purpose alone!

Please do share problems and concepts in this board, and ask uncountable number of doubts. Also discuss about books which can be helpful for RMO preparation. Some of them I recommend are :

• Challenge and Thrills of Pre College Mathematics

• Problem Solving Strategies by Arthur Engel

• RMO and INMO book of Arihant Publication by Rajeev Manocha

Miscellaneous

Please do share Concepts of the Day and also the problems related to it. Do link question papers so that all of us can do them together. I hope the members of our community would be able to represent their respective countries in the IMO!

Note by Swapnil Das
1 year, 8 months ago

## Comments

Sort by:

Top Newest

Great step!

Well, there are many who would be interested in this discussion.The toughest I feel is the number theory part of RMO. What are some of the good sources to prepare for it.

There are some exceptionally brilliant people on Brilliant who have the experience of RMO,INMO,IMOTC and IMO.It would be interesting if they take part in this discussion. · 1 year, 8 months ago

Log in to reply

Number Theory by Burton is the best for preparing for RMO's number theory part, as I think. · 1 year, 7 months ago

Log in to reply

Excellent job! Keep on posting RMO type problems!

@Shivam Jadhav · 1 year, 8 months ago

Log in to reply

@Shivam Jadhav I appreciate your efforts of posting RMO problems. It would have been great if you posted proof problems also as note. (Like Xuming does for geometry). Thanks anyways for your step :) · 1 year, 8 months ago

Log in to reply

Surely Nihar · 1 year, 8 months ago

Log in to reply

Hii I am also preparing for RMO can you tell me topics or chapters(syllabus) which we have to prepare for RMO... · 1 year, 8 months ago

Log in to reply

The major chapters are Number Theory, Algebra, Geometry and Combinatorics · 1 year, 7 months ago

Log in to reply

Yes, I will provide you the complete Syllabus in few hours☺ · 1 year, 8 months ago

Log in to reply

There's no as such particular syllabus for RMO. · 1 year, 8 months ago

Log in to reply

Sir which book would u suggest for number theory · 1 year, 7 months ago

Log in to reply

Hello sir, I have been introduced to David Burton's Number theory, which I have started using. I would recommend you the same. · 1 year, 7 months ago

Log in to reply

Sir firstly i would like to appreciate this step of yours of making rmo board thank you :) and also thank you for your suggestion · 1 year, 7 months ago

Log in to reply

Welcome , it was my pleasure benefiting you :) · 1 year, 7 months ago

Log in to reply

@Swapnil Das Don't you think it would be better if you create a second part of this note? That way it will be better for people to see comments more easily and respond to them since there are so many comments in this note already. · 1 year, 8 months ago

Log in to reply

I liked you suggestion and I have created a new thread.Thanks. · 1 year, 8 months ago

Log in to reply

Thank you so much for your efforts. · 1 year, 8 months ago

Log in to reply

No need to thank me. This is our continued effort :) · 1 year, 8 months ago

Log in to reply

Comment deleted Sep 26, 2015

Log in to reply

No I don't think this will do good. · 1 year, 8 months ago

Log in to reply

ok · 1 year, 8 months ago

Log in to reply

If $$p$$ is a prime number, then prove that $$7p + 3p -4$$ is not a perfect square. · 1 year, 8 months ago

Log in to reply

If it is $$7p+3p-4=10p-4$$ then it is extremely trivial. Applying the same logic as in the answer below,

A]$$p\equiv 1 (mod 4)$$: $$10p-4\equiv6 \equiv 2 (mod 4)$$

B] $$p\equiv -1 (mod 4)$$ : $$10p-4\equiv -14 \equiv 2 (mod 4)$$

So neither of the two give us $$\equiv 0,1 (mod 4)$$.

Therefore, there are no such square except for $$p=2$$ · 1 year, 8 months ago

Log in to reply

Yep. This one is extremely easy. · 1 year, 8 months ago

Log in to reply

If its $$7{p}^{2}+3p-4$$ then,

It can be proved easily for $$p=2$$ .

All perfect squares are $$\equiv 1,0 (mod 4)$$

We know that for all $$p$$ excluding $$2$$, $${p}^{2}\equiv 1 (mod 4)$$

As all primes are odd numbers, we can segregate the primes into two cases:

A] $$p\equiv 1 (mod 4)$$ :

For this, $$7{p}^{2}+3p-4\equiv 7*1+3*1-4\equiv 6\equiv 2 (mod 4)$$ Therefore this case has no squares formed.

B] $$p\equiv -1 (mod 4)$$:

For this, $$7{p}^{2}+3p-4\equiv 7*1+3*(-1)-4\equiv 0 (mod 4)$$

This case seems to satisfy the required condition.

For this we need to apply $$(mod 11)$$. All squares are $$\equiv 0,1,3,4,5,9 (mod 11)$$. This can be proved.

So we just need to check that $$1,3,4,5,9 (mod 11)$$ is not satisfied for any $$p\equiv 1,3,5,7,9 (mod 11)$$ in the equation $$7{p}^{2}+3p-4$$.

For squares $$\equiv 0 (mod 11)$$, We just need to check for $$p=11$$ and we will find out that this neither gives us a square.

Therefore there are no squares of the form $$7{p}^{2}+3p-4$$ · 1 year, 8 months ago

Log in to reply

Is it $$7{p}^{2}+3p-4$$? · 1 year, 8 months ago

Log in to reply

Must be. · 1 year, 8 months ago

Log in to reply

try this...

Let $$a$$ be positive real number such that $$a^3 = 6(a + 1)$$ then prove that $$x^2 + ax + a^2 - 6 = 0$$ has no real roots. · 1 year, 8 months ago

Log in to reply

done · 1 year, 8 months ago

Log in to reply

show · 1 year, 8 months ago

Log in to reply

$${a}^{3}-6a-6=0$$

Let $$a=b+2/b$$

Therefore, $${ \left( b+2/b \right) }^{ 3 }-6(b+2/b)-6=0$$

Simplifying we get that, $${b}^{6}-6{b}^{3}+8=0$$

Therefore, $${b}^3=4$$ or $$2$$

Substitute these values to get $$a$$.

That time we see that only one real solution of $$a$$ occurs which is, $$a={2}^{1/3}+{2}^{2/3}$$

We see that, $${a}^{2}-6=6/a$$

Substituting this value in $${x}^{2}+ax+{a}^{2}-6=0$$ we get that,

$$a{x}^{2}+{a}^{2}x+6=0$$

Assume that the roots of these quadratic equation are real, Then using formula for roots for quadratic equations,

$$x=\frac { -a\pm \sqrt { { a }^{ 2 }-24 } }{ 2 }$$

Then substituting the acquired value of $$a$$ in this equation we get that $$x$$ is a complex number. Hence, our assumption was wrong.

Hence proved that roots of the given quadratic equations are not real. · 1 year, 8 months ago

Log in to reply

This can also be done using Cardano's method of finding solutions of a cubic equation. · 1 year, 1 month ago

Log in to reply

Can you explain why you took the initial substitution of a= b+ (2/b) ?

Thanks in advance! :-)

Ingenious solution nonetheless! · 1 year, 1 month ago

Log in to reply

You're a genius. _/_ · 1 year, 8 months ago

Log in to reply

Nah. Just able to solve RMO problems · 1 year, 8 months ago

Log in to reply

Solving RMO probs is not easy bro ;) · 1 year, 8 months ago

Log in to reply

Probably yes · 1 year, 8 months ago

Log in to reply

nice... Try my another question which i am going to post · 1 year, 8 months ago

Log in to reply

Correct! · 1 year, 8 months ago

Log in to reply

Thanks! · 1 year, 8 months ago

Log in to reply

Can someone suggest a good book for combinatorics with lots of examples and problems with solutions for RMO · 1 year, 5 months ago

Log in to reply

Guys , looking for varied solutions here.

Instead of posting questions here , we will post them as a note and give their respective links here. Is this okay? · 1 year, 8 months ago

Log in to reply

ok!! · 1 year, 8 months ago

Log in to reply

Do you have to be in Romania in order to qualify for RMO? · 1 year, 8 months ago

Log in to reply

No, it is the board of the Indian RMO. · 1 year, 8 months ago

Log in to reply

Oh sorry, wrong RMO. · 1 year, 8 months ago

Log in to reply

Has anybody heard of CHINESE DUMBASS NOTATION. (LOL) But keeping the name aside, its a very good tool for solving most of the types of inequalities in RMO. Read about this!! It is very helpful. · 1 year, 8 months ago

Log in to reply

Yes , I have heard about it. But I have never applied it though :P · 1 year, 8 months ago

Log in to reply

Ok XD · 1 year, 8 months ago

Log in to reply

Comment deleted Dec 18, 2015

Log in to reply

Wow Saarthak I didn't know you qualified RMO last year. Great achievement. I thought that u too like me had just cleared pre -rmo · 1 year, 5 months ago

Log in to reply

Is it closed now? A second question : Is $$GEOMETRY$$ banned here? Not a single stuff.... · 1 month, 3 weeks ago

Log in to reply

Today, i got a call from rmo office, and they were saying that from my region only 3 student filled the form. They cant conduct exam on the preferred center by me. And they were saying i had to come to jaipur(capital)... -_- · 1 year, 6 months ago

Log in to reply

Ohh , In which city and state do you live? · 1 year, 6 months ago

Log in to reply

I live in Nohar (northern rajasthan).. · 1 year, 6 months ago

Log in to reply

Lol....It looks like you are already selected. · 1 year, 6 months ago

Log in to reply

How can a person get selected without giving the exam? Weird... · 1 year, 6 months ago

Log in to reply

No I meant that in any region (at least for Delhi 17 are selected) at least 20 are selected. So even getting low marks you have a high probability for selection.

Ps-it was a joke and I didn't mean he is already selected. · 1 year, 6 months ago

Log in to reply

how? · 1 year, 6 months ago

Log in to reply

@Calvin Lin Sir, is it possible for you to close this note since we already have a part 2 for this note. · 1 year, 7 months ago

Log in to reply

I don't see why this note should be locked. Staff · 1 year, 7 months ago

Log in to reply

Please sir, don't lock this note. I'm still benefiting from it. · 1 year, 7 months ago

Log in to reply

@Swapnil Das @Mehul Arora @Dev Sharma Check out this link http://artofproblemsolving.com/community/c3176indiacontests.It contains the problems of several contests held in India (including RMO,INMO and problems for the IMOTC). · 1 year, 8 months ago

Log in to reply

Thanks! @Svatejas Shivakumar · 1 year, 8 months ago

Log in to reply

Log in to reply

Comment deleted Sep 17, 2015

Log in to reply

RMO Set · 1 year, 8 months ago

Log in to reply

CHECK THIS OUT INEQUALITY. · 1 year, 8 months ago

Log in to reply

Hi,

• So what is the Theorem of the day?

• Any new topic?

· 1 year, 8 months ago

Log in to reply

Chinese Remainder Theorem!!! ,would be the best · 1 year, 8 months ago

Log in to reply

OK, the topic of the day from my side is:

Euler's Theorem · 1 year, 8 months ago

Log in to reply

Cauchy-Schwartz Inequality · 1 year, 8 months ago

Log in to reply

can students of class xii participate in RMO? · 1 year, 8 months ago

Log in to reply

Not really. · 1 year, 8 months ago

Log in to reply

are you sure, sir? · 1 year, 8 months ago

Log in to reply

Yes, Sir. 12th grade is now restricted to appear RMO. · 1 year, 8 months ago

Log in to reply

Guys, if you want to solve a RMO problem, see this one https://brilliant.org/problems/a-geometry-problem-by-saarthak-marathe-2/?group=Z7UjgQAVmgvN . For more,see my sets. · 1 year, 8 months ago

Log in to reply

I have posted the note · 1 year, 8 months ago

Log in to reply

I have a doubt:

Find the sum of the squares of the roots of the equation :

$${ x }^{ 2 }+7[x]+5=0$$ · 1 year, 8 months ago

Log in to reply

Comment deleted Sep 18, 2015

Log in to reply

Comment deleted Sep 17, 2015

Log in to reply

He is wrong, they aren't integer solutions and he assumed in the beginning that x is an integer. Morover, it should have been 39. · 1 year, 8 months ago

Log in to reply

It should be 39 is case 1. And the values you are getting are not integers so do you think those solutions are valid? · 1 year, 8 months ago

Log in to reply

Very close to the answer, I will tell you today. · 1 year, 8 months ago

Log in to reply

Is this mod x? If it is then there are no solutions to this equation. · 1 year, 8 months ago

Log in to reply

No, it is the ceiling function. · 1 year, 8 months ago

Log in to reply

Why did you delete my comment? If it is the ceiling function then the answer is 92 and if it is the floor function then the answer is 95. · 1 year, 8 months ago

Log in to reply

Really, that means it is the smallest integer function, this question becomes easier · 1 year, 8 months ago

Log in to reply

It is floor function · 1 year, 8 months ago

Log in to reply

No. It is greatest integer function, which means greatest integer less than the given number. For example, $$[3.23423]=3$$

$$[-4.243252]=-5]$$ · 1 year, 8 months ago

Log in to reply

I know what is greatest integer function, but its notation is |_| is like this as it is the floor function. · 1 year, 8 months ago

Log in to reply

greatest integer function can be called as a floor function. · 1 year, 8 months ago

Log in to reply

That's what I said didn't I. I said it has the same notation because it is the floor function. · 1 year, 8 months ago

Log in to reply

Is the answer -7 (by any chance) ? · 1 year, 8 months ago

Log in to reply

How can sum of squares be negative? · 1 year, 8 months ago

Log in to reply

Oh..well--I think I overlooked the word "squares"..x'tremely sorry!! · 1 year, 8 months ago

Log in to reply

OK, so the topic of the Day, from my side, is :

$$\huge\ Vieta's Formula$$ · 1 year, 8 months ago

Log in to reply

Can you give the links from where you found them? · 1 year, 8 months ago

Log in to reply

The question? · 1 year, 8 months ago

Log in to reply

Yeah the question or if you can find wiki's. Because there are many names of theorem which I don't know but wjen I see them, they are actually quite often used by me · 1 year, 8 months ago

Log in to reply

inequalities · 1 year, 8 months ago

Log in to reply

OK,good idea! Even I haven't started that topic😛 · 1 year, 8 months ago

Log in to reply

Can anyone share Topic of the day, so that we get to study it, and do some problems on it? · 1 year, 8 months ago

Log in to reply

We must start with inequalities · 1 year, 8 months ago

Log in to reply

Please someone tell me good brilliant questions that are good for RMO preparation except Shivam Jadhav's problems. · 1 year, 8 months ago

Log in to reply

Try the set, " Openly welcome for future Mathematicians". · 1 year, 8 months ago

Log in to reply

what about RMO forms? · 1 year, 8 months ago

Log in to reply

The forms for some of the regions have already been uploaded. In which region are you giving RMO? · 1 year, 8 months ago

Log in to reply

Can you elaborate? · 1 year, 8 months ago

Log in to reply

i am asking about the date when forms would be available · 1 year, 8 months ago

Log in to reply

do you know????? · 1 year, 8 months ago

Log in to reply

You can write in any of the regions(as per your convenience). See this link for the list of regions.http://olympiads.hbcse.tifr.res.in/enrollment/list-of-rmo-coordinators.Note that a region may be further divided into sub regions. You may see the website for your region or contact your regional coordinator for more details. · 1 year, 8 months ago

Log in to reply

i live in hanumangarh district in rajasthan · 1 year, 8 months ago

Log in to reply

I don't know about Rajasthan region much.You can contact your regional coordinator from the link above. · 1 year, 8 months ago

Log in to reply

Here, Pre RMO is kinda integer type exam. No proving😜 · 1 year, 8 months ago

Log in to reply

No pre RMO in my region 😟 · 1 year, 8 months ago

Log in to reply

@Swapnil Das don't you think rmo is more of higher thinking with concept. Only concept is not what all it requires. · 1 year, 8 months ago

Log in to reply

Think of finding the Area of triangle without knowing the formula. Concept is the very fist thing to be cleared. After knowing varied concepts, brain works better and you can think stuff in a number of ways and directions. · 1 year, 8 months ago

Log in to reply

Yes , It requires out of box thinking too... · 1 year, 8 months ago

Log in to reply

OMGOMGOMGOMGOMGOMGOMGOMGOMG!!!!!!! · 1 year, 8 months ago

Log in to reply

Delete the comment. · 1 year, 8 months ago

Log in to reply

Don't worry , the more he comments , the sooner his account will be deleted and he will be banned :) · 1 year, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...