Post as many problems as you can for others to try.

Please don't post problems from past year RMO papers.

Have FUN!

Here is a problem to start with :

Let \(a, b, c, d, e, f\) be real numbers such that the polynomial equation

\[x^{8} - 4x^{7} +7x^{6} +ax^{5}+ bx^{4} + cx^{3} + dx^{2} + ex + f = 0\]

has eight positive real roots. Determine all possible values of \(f\)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestSince we were given that the roots were positive reals,we get motivated and tempted to use,\(R.M.S\geq A.M\geq G.M\geq H.M\),also since we have been given,\(\sum \alpha\) and \(\sum \alpha\beta\) we try to find the value of \(\sum \alpha^2=4^2-2\times 7=2\).We apply \(R.M.S \geq A.M\) \[\sqrt{\dfrac{2}{8}} \geq \dfrac{4}{8}\\ \Longrightarrow \dfrac{1}{2} \geq \dfrac{1}{2}\],hence equality occurs,meaning that all the roots are equal.Now,that means,\(A.M=G.M\),hence\[\dfrac{4}{8}=\sqrt[8]{f}\\ \Longrightarrow \dfrac{1}{256}=f\],hence there is only one value of \(f=\dfrac{1}{256}\).Is this correct Harsh?

Log in to reply

In your solution you need prove that \[(\sum_{cyclic}ω)^{2}-\sum_{cyclic}ω_{1}ω_{2}=\sum_{cyclic}ω^{2}\]

Log in to reply

That's an identity that is true for all numbers, real and imaginary. And there should be a 2 infront of that second summation.

Log in to reply

Log in to reply

If \(x,y,z\) are positive real numbers, prove that,

\[\large (x+y+z)^2 (yz+zx+xy)^2 \le 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)\]

Give me as much different solutions please .

Log in to reply

U may use the concept of fermat pt to solve this one..then it will be very easy...suppose three line segments AF,BF,CF of length x,y,z meet each other at120 degrees and now use cosine rule to find out the lengths of the triangle and see that each term on the rhs of the inequality is actually squares of the sides...then express xy+yz+zx as \(\frac{4}{\sqrt{3}}T\) where T is the area of the triangle....then use \(T=\frac{abc}{4R}\) and see that the inequality reduces to 3R greater than or equal to (x+y+z) which is obvious since F is the fermat pt

Log in to reply

You have a set of consecutive positive integers,\(\{1,2,3,...,199,200\}\),if you choose \(101\) numbers from these at random,then prove that there at-least two numbers of which one divides the other.

Log in to reply

If \(x\) and \(y\) are integers such that \(y^2+3(xy)^2=30x^2+517\), Prove that \(3(xy)^2=588\)

Log in to reply

Let \(y^{2} = b\) and \(x^{2} = a\).

\(b(1 + 3a) = 30a +517 \)

\(\implies b(1+3a) -10(3a + 1) = 507 \)

\(\implies (b-10)(3a+1) = 507 \times 1 = 13 \times 39 = 3 \times 169\)

Since a , b are integers, after checking through all 6 cases, we get only admissible values of \((x ,y) = ( \pm 2, \pm 7)\).

This forces \(3(xy)^{2}= 588\).

\(\QED \)

Log in to reply

Find all positive integer solutions \(a\), \(b\) and \(c\) such that

\[\dfrac {7}{a^2} + \dfrac {11}{b^2} = \dfrac {c}{77}.\]

Log in to reply

\(\text{THIS IS ONLY AN OUTLINE OF THE SOLUTION, NOT A COMPLETE SOLUTION.}\)

\[\dfrac {7 }{a^2} + \dfrac {11 }{b^2} = \dfrac{c}{77}.\]

Case 1:\(c>1\)This forces \(a^{2} < 7 \text{ and } b^{2} < 11\)

Now it is easy to check that \( (a,b,c) = (1,1,1386) \text{ and } (a,b,c) = (3,3,154)\) satisfy the given conditions.

Case 2:\(c \leq 77\)Since \(c \leq 77\) this implies \(\dfrac{c}{77} \leq 1\).

Let \(\dfrac{c}{77} = \dfrac{1}{k}\) for some integer \(k \geq 1\).

\(\implies c = \dfrac{77}{k}\)

Now since \(c\) is a positive integer , this forces

\(k \text{ | } 77\).Possible values for \(k = 1,7,11,77\).

Now plugging the possible values of \(c\) in the given equation and using Simon's Favourite Factoring Trick, we can easily solve for \(a,b\)

Log in to reply

Find all primes \(p\) such that for all prime \(q\), the remainder of \(p\) upon divison by \(q\) is squarefree.

Log in to reply

I think that 3 is the only prime number. I used the fact that all primes>3 is of the form \(\pm 1 \pmod{6}\).

Log in to reply

x, y, z are positive real numbers such that x^2+y^2+z^2=3 Prove that x+y+z >= (xy)^2+(yz)^2+(zx)^2

Sorry for not using latex as I was in a hurry.

Log in to reply

Can u send the solution of this question ? I have been trying. No progress ....

Log in to reply

I didn't get it.

Log in to reply

Let \(x,y\) and \(z\) be positive real numbers such that \(x+y+z=1\). Prove that \(\frac{yz}{1+x}+\frac{zx}{1+y}+\frac{xy}{1+z} \le \frac{1}{4}\).

Log in to reply

I got this. This is simple AM-HM inequality. Write 1+x=(1-y)+(1-z). Similarly for others

Apply AM-HM and then manipulate on the RHS.

Log in to reply

Find all reals \(x, y, z\) in (1,∞) such that

\[x + y + z + \dfrac{3}{x-1}+ \dfrac{3}{y-1} + \dfrac{3}{z-1} = 2(\sqrt{x+2} + \sqrt{y+ 2} + \sqrt{z+2})\]

Log in to reply

https://brilliant.org/problems/three-variables-one-equation/?group=4bQcwX9pLM2R#!/solution-comments/106741/

Log in to reply

If \(a,b,c\) are positive integers such that \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}<1\). Prove that the maximum value of \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) is \(\dfrac{41}{42}\)

Log in to reply

If none of \(a,b,c=2\), the maximum value of the expression will be \(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}<\frac{41}{42}\). So, one of them has to \(2\) say \(a\). So we have to prove that the maximum value of \(\frac{1}{b}+\frac{1}{c}\) is \(\frac{10}{21}\). If none of \(b,c=3\), the maximum of the expression will be \(\frac{1}{4}+\frac{1}{5}=\frac{9}{20}<\frac{10}{21}\). So, one them has to be \(3\) say \(b\). Therefore, \(\frac{1}{c} \le \frac{1}{7}\).Therefore the maximum value of \(c\) is \(7\) and the maximum value of \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) is \(\frac{41}{42}\).

Not sure if this is the best method.

Log in to reply

Find all primes \(x,y\) such that \(x^{y}-y^{x}=xy^{2}-19\).Enjoy solving this.

Log in to reply

Using FLT, we have

\[x^y \equiv x \pmod{y} \Rightarrow x \equiv -19 \pmod {y} \Rightarrow x+19 \equiv 0 \pmod{y}\]

\[y^x \equiv y \pmod{x} \Rightarrow -y \equiv -19 \pmod {x} \Rightarrow 19-y \equiv 0 \pmod{x}\]

From here, it is clear that either \(x=2\) or \(y=2\). We now separate this into 2 cases:

Case 1:\(y=2\)When \(y=2\), we have \(19 - 2 = 17 \equiv 0 \pmod{x}\), which implies \(x=17\). The solution in this case is \((17, 2)\). Checking, however, we find that this clearly doesn't work. So, there are no solutions in this case.

Case 2:\(x=2\)When \(x=2\), we have \(19+2 = 21 \equiv 0 \pmod {y}\), which implies \(y=3\) or \(y=7\). Checking, we find both these solutions work. Thus, 2 solutions exist in this case: \((2, 3)\) and \((2,7)\).

Therefore, only 2 solutions exist: \((2, 3)\) and \((2,7)\).

Log in to reply

would you please explain how x = -19 mody

Log in to reply

Log in to reply

Using FLT and modulo \(x\) and \(y\) we get that,\(x+19 \equiv 0 \pmod{y},y-19 \equiv 0 \pmod{x}\),now just use parity(as suggested by @Sharky Kesa).

Log in to reply

Let \(p\) and \(q\) be distinct primes. Show that

\(\displaystyle \sum_{j=1}^{(p-1)/2}[\frac{qj}{p}]+ \sum_{j=1}^{(q-1)/2}[\frac{pj}{q}]=\dfrac{(p-1)(q-1)}{4}\).

Note that \([ \ ]\) represents the greatest integer function.

Log in to reply

Let \(S={(x,y)|x,y \in N, \quad 1 \le x \le (p-1)/2, \quad 1 \le y \le (q-1)/2}\).

We note that the set \(S\) can be partitioned into two sets \(S_1\) and \(S_2\) according as \(qx>py\) or \(qx<py\). Note that there are no pairs in such \(S\) such that \(qx=py\).

The set \(S_1\) can be described as the set of all pairs \((x,y)\) satisfying \(1 \le x \le (p-1)/2, \quad 1 \le y < qx/p\). Then \(S_1\) has \(\displaystyle \sum_{j=1}^{(p-1)/2}{\frac{qj}{p}}\) elements.

Similarly, \(S_2\) has \(\displaystyle \sum_{i=1}^{(q-1)/2}{\frac{pi}{q}}\). Hence, we get the required result.

Log in to reply

Let \(a,b,c,d \in \mathbb{N}\) such that \(a \ge b \ge c \ge d \) and \(d \neq 0\). Show that the equation \(x^4 - ax^3 - bx^2 - cx -d = 0\) has no integer solution in \(x\).

Please post a "

complete solution" like you would do in RMO.Log in to reply

I'll have a crack at it. As \(d \neq 0\), \(x \neq 0\).

Denote \(y = \frac{1}{x}\).

We have

\[x^4 = ax^3 + bx^2 + cx + d \quad \quad\]

\[1 = ay + by^2 + cy^3 + dy^4 \quad \quad (1)\]

Now, we separate this into 2 cases:

Case 1:\(x = -n\) where \(n \in \mathbb{N}\), then \(y=-\dfrac{1}{n}\). Substituting this into \((1)\), we get

\[\dfrac {-a}{n} + \dfrac {b}{n^2} + \dfrac {-c}{n^3} + \dfrac {d}{n^4} = 1\]

\[\dfrac {-an+b}{n^2}+\dfrac{-cn+d}{n^4}=1\]

Here, the LHS is non-positive since \(a \geq b\) and \(c \geq d\), whereas the RHS is positive. Thus no solutions exist in this case.

Case 2:\(x = n\) where \(n \in \mathbb{N}\), then \(y=\dfrac{1}{n}\). Substituting this into \((1)\), we get

\[\dfrac {a}{n} + \dfrac {b}{n^2} + \dfrac {c}{n^3} + \dfrac {d}{n^4} = 1\]

This implies that \(n \neq 1\), and

\[\dfrac {a}{n} < 1\]

\[1 \leq \dfrac {a}{n} + \dfrac {a}{n^2} + \dfrac {a}{n^3} + \dfrac {a}{n^4} < \displaystyle \sum_{i=1}^{\infty} \dfrac{a}{n^i} = \dfrac {a}{n-1}\]

These two equations imply that

\[\dfrac {n}{a} > 1\]

\[\dfrac {n-1}{a} < 1\]

There are no such \(n\) that satisfy the above two inequalities. Thus, no solutions exist.

Log in to reply

Isn't this from INMO 2013 ??

Log in to reply

Yes it is.

Log in to reply

Log in to reply

Let \(a,b,c\) be positive reals satisfying \((a+b)(b+c)(c+a) = 1\).

Prove that \(ab+bc+ca \leq \frac{3}{4}\)

Log in to reply

Using am gm

(a + b) + (b + c) + (c + a) > 3

a + b + c > 3/2

now using cauchy,

\(3(a^2 + b^2 + c^2) > (a + b + c)^2\)

so

\(a^2 + b^2 + c^2 > 3/4\)

using cauchy again,

\((a^2 + b^2 + c^2)^2 > (ab + bc + ca)^2\)

3/4 > ab + bc + ca

Log in to reply

Nice solution. You could also have used the fact that \((a+b+c)^{2} \ge 3(ab+bc+ca)\)

Log in to reply

Dev,your solution is incorrect. Check the solution again,you will find your flaw.

Log in to reply

You got the following two inequalities,\[a^2+b^2+c^2>ab+bc+ca\\ and\\ a^2+b^2+c^2>\dfrac{3}{3}\],what next?How can you conclude that,\[\dfrac{3}{4}>ab+bc+ca\]?

Log in to reply

(sum(a^2))(sum(b^2)) > (sum(ab))^2

sum is cyclic

Log in to reply

If \(a,b \ge 0\). Prove that \(\sqrt{2}(\sqrt{a(a+b)^{3}}+b \sqrt{a^{2}+b^{2}}) \le 3(a^{2}+b^{2})\).

Log in to reply

Prove that

\[ 2 < (\dfrac{n+2}{n+1}) ^ {(n+1)}\]for all integers \(n > 0\).Log in to reply

Well its easy just use Bernoulli's inequality that (1+x)^n > 1+nx

Log in to reply

Let \(p>3\) be a prime number. Suppose \(\displaystyle \sum_{k=1}^{p-1}{\frac{1}{k}}=\frac{a}{b}\) where \(\gcd(a,b)=1\). Prove that \(a\) is divisible by \(p^{2}\).

Log in to reply

I am able to prove that a is divisible by p. But not able to prove divisible by p^2. Could u send the solution ?

Log in to reply

Please post the proof, thanks!

Log in to reply

Log in to reply

Log in to reply

Log in to reply

I have the solution only till \(a\) is divisible by \(p\). To prove that \(a\) is divisible by \(p^{2}\) is given below the solution as a challenge.

Log in to reply

In a triangle \(ABC\) , D is the midpoint of side AB and E is the point of trisection of side BC nearer to C.

Given that \(\angle ADC = \angle BAE \), find \(\angle BAC\).

Log in to reply

Is the answer angle BAC = 90 degrees ? I hope so My proof : Apply Menelaus theorem on triangle BCD with transversal EA. Obtain that F is the midpoint of CD. Then its simple angle chasing :)

Log in to reply

Shrihari,we get that \(CF=2DF\)

Log in to reply

Log in to reply

(CF/FD)(DA/AB)=1Log in to reply

Log in to reply

Yup I also got 90 degrees.

Log in to reply

Yes , Menelaus is indeed a good approach for this problem. For shortening the solution: Once you get F is the midpoint of CD , you can directly say that \(\angle BAC = 90^\circ\) , do you see why?

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

There are given \(2^{500}\) points on a circle labeled \(1, 2, . . . , 2^{500}\)in some order. Prove that one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums of the pairs of numbers at the endpoints of the chosen chords are equal.

Log in to reply

Given seven arbitrary distinct real numbers, show that there exist two numbers \(x\) and \(y\) such thst \(0<\dfrac{x-y}{1+xy}<\dfrac{1}{\sqrt{3}}\).

Log in to reply

The question is easy. Consider x and y to be some tan(A) and tan(B) respectively and consider A and B to vary from 0 to pi so that all reals are covered. now divide it into six equal portions spaced by pi/6. Now u have 6 pigeons and 7 holes and you are done. Then that expression is simply tan(A-B).

Log in to reply

If a circle goes through point \(A\) of a parallelogram \(ABCD\), cuts the two sides \(AB,AD\) and the diagonal \(AC\) at points \(P,R\) and \(Q\) respectively. Prove that \((AP×AB)+(AR×AD)=AQ×AC\).

Log in to reply

I hope you all have drawn the diagram.(Consider all the segments below as vectors in the given direction.)

Let \(AS\) be the diameter of given circle.

\(AB+AD=AC\)

Therefore taking dot product with\(AS\)

\(AB.AS+AD.AS=AC.AS \)

Using, \(a.b=|a|*|b|*cos(\theta) \) [\( \theta\) is angle between the vectors]

\(|AB|*|AP|+|AR|*|AD|=|AQ|*|AC| \)

Log in to reply

Let \(ABC\) be a triangle and \(P\) be a point inside it.Let \(x,y,z\) denote the lengths of \(PA\),\(PB\),\(PC\).Let \(a,b,c\) be the lengths of sides \(BC\),\(CA\),\(AB\).Find all points \(P\) such that \(axy+byz+czx=abc\).

Log in to reply

Find the least positive integer n such that 2549 divides \(n^{2545} - 2541\)

Log in to reply

Please post the solution , Thanks!

Log in to reply

Let \(ABC\) be an acute-angled tiangle .Let O denote its circumcenter.Let \(\Pi\) be the circle passing through the points A,O,B.Lines CA and CB meet \(\Pi\) again at P and Q respectively.Prove that PQ is perpendicular to the line CO.

Log in to reply

Harsh,could you please add a figure?

Log in to reply

Hint:Let \(\overrightarrow{QP} \cap \overleftrightarrow{CO} = \{R\}\) and \(\overleftrightarrow{CO} \cap \Pi = \{S\}\) and then it will suffice to prove that \(\Delta CRP \sim \Delta CAS\)Log in to reply

An easy INMO geometry problem:

Let \(ABC\) be a triangle with circumcircle \(\Gamma\). Let \(M\) be a point in the interior of triangle \( ABC\) which is also on the bisector of \(\angle A\). Let \(AM, BM, CM\) meet \( \Gamma\) in \( A_{1}, B_{1}, C_{1}\) respectively. Suppose \(P\) is the point of intersection of \( A_{1}C_{1}\) with \(AB\) and \(Q\) is the point of intersection of \(A_{1}B_{1}\) with \(AC\). Prove that \(PQ\) is parallel to \(BC\) .

Log in to reply

Let \(a,b \in R, a \neq 0\). Show that \(a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{b}{a} \ge \sqrt{3}\).

Log in to reply

Well the question becomes very easy if u make it a quadratic in b and then prove that the Discriminant is less than 0. I did it that way.

Log in to reply

If \(a,b,c,d,e\) are real numbers, prove that the roots of \(x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e=0\) cannot all be real if \(2a^{2}<5b\).

Log in to reply

Let roots be \(p,q,r,s,t\).

Sum of roots = -a & Product of roots taken two at a time = b.

Therefore Sum of squares of roots = \(a^{2} - 2b\)

If all roots are real, then by Titu's Lemma,

\(a^{2} -2b \geq \frac{a^{2}}{5} \)

\(\implies 2a^{2} \geq 5b\)

Therefore the polynomial cannot have all roots real if \( 5b > 2a^{2}\).

Log in to reply

Show that \(n^{5}+n^{4}+1\) is not prime for \(n>1\).

Log in to reply

\(n^{2} + n + 1\) is factor of the given expression.Hence the result is obvious.

Log in to reply

Prove that

\(1<\frac { 1 }{ 1001 } +\frac { 1 }{ 1002 } +.........+\frac { 1 }{ 3001 } <\frac { 4 }{ 3 } \)

Log in to reply

It is easy. I got a calculus and a non calculus answer

Log in to reply

Since Rmo is a precalculus olympiad, answer without calculus will be better:P

Log in to reply

Log in to reply

Log in to reply

Alternate text. Type the text which you wish to be displayed if the image doesn't load in the third brackets, and paste the image URL in the second brackets. In the following example, the image URL is http://s12.postimg.org/5qhlgca71/untitled.png, and I wrote the code This is an image file :D. Here's how it appears:

First use an image uploading website (I prefer postimg.org ), upload your picture, and copy the image link. The format for uploading images in a solution isthis is a copy paste from this note on moderators

Log in to reply

If \(m\) is a positive real number and satisfies the equation \[2m=1+\frac{1}{m}+\sqrt{m^{2}-\frac{1}{4}}\] Find \(m\).

Log in to reply

Could you send the solution of this question. I tried expanding and landed up with a fourth degree polynomial whose roots I am unable to guess

Log in to reply

Find all pairs \((m,n)\) such that \(2^m+3^n\) is a perfect square. (A popular NT problem)

Log in to reply

Its easy.Only one solution (2,2)

Log in to reply

Wrong. Check again.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Let d(k) denote the sum of the digits of k in base10. Find all natural numbers n such that n, d(n), d(d(n)) sum to 2015.

Log in to reply

I am getting no solution. Is that right ? My proof : n,d(n),d(d(n)) are congruent to the same number say 'x' (mod 9). So their sum is congruent to 3x (mod 9). Therefore 2015 is congruent to 3x ( mod 9) which is impossible as 3 does not divide 2015

Log in to reply

Here is a geometry question. Waiting for numerous solutions. In a triangle ABC, excircle opposite to A touches BC at point Q. Let M be the midpoint of BC. Incircle of triangle ABC touches BC at point P. Prove that PM=MQ. I have been trying this question but I am stuck. So please solve this question.

Log in to reply

Its easy , note that \(BP=QC=s-b\) where \(s\) is the semiperimeter of the triangle and \(b\) is length of side opposite to \( \angle B\) respectively. Since \(M\) is mid point of \(BC\) , the result follows.

Log in to reply

Are protractors allowed for RMO? Nothing about it is mentioned in the website and the hall ticket.

Log in to reply

Log in to reply

Oh nice !

Log in to reply

Try this set guys.

Log in to reply

\(1991^{k}|(1990^{1991^{1992}}+1992^{1991^{1990}})\)

\(max(k)=?\)

Details:Log in to reply

1991

Log in to reply

How ? Could u give some rough outline of your solution ?

Log in to reply

https://brilliant.org/problems/triangular-grid-problem/?group=U3Rx70QSpE0z&ref_id=1044906

Log in to reply

Log in to reply

Yes..... Can you post solution ?

Log in to reply

Everyone here who have posted questions on this board and have not got any solutions please post the solutions now as RMO is only 42 hours from now. Its better that we understand the methods. Sorry that I cannot send the solution of my geometry question as I myself have been trying that question but could not arrive at the solution.

And please post some more questions if possible :)

Log in to reply

Done :)

Log in to reply

Thanks

Log in to reply

Let \(x,y\) be real numbers such that \(x,y > 1\)

Prove that \(\dfrac{x^{2}}{y-1} + \dfrac{y^{2}}{ x-1} \geq 8\)

Log in to reply

Just use AM-GM inequality directly on the expression in the LHS and then use the fact that (a-2)^2>0 Then rearrange a bit. You get the required answer

Log in to reply

Alternatively we can substitute p = x-1 and q = y-1.

Then it will be easy manipulate the expression.

Log in to reply

Nice idea!

Log in to reply

Since the expression is symmetric, let's assume without loss of generality that \(x \ge y\).

\(\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge \frac{y^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 4+\frac{y^{2}}{x-1} \implies \frac{x^{2}}{y-1} \ge 4\). Minimum value of \(\frac{x^{2}}{y-1}\) is \(4\) and it occurs if and only if \(x=y=2\).

Therefore, \(\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 8\).

Log in to reply

Hmm you have not proven that min value of \(\frac{x^{2}}{y-1}\) is 4.

Log in to reply

Log in to reply

Sorry didn't see this post but since I found it, here you go - 65 bugs are placed at different squares of a 9*9 square board. Abug in each moves to a horizontal or vertical adjacent square. No bug makes 2 horizontal or vertical moves in succession. Show that after some moves, there will be at least 2 bugs in the same square.

Log in to reply

\(a,b,c\) are real numbers such that \(a+b+c=0\) and \(a^2+b^2+c^2=1\). Prove that \(a^{2}b^{2}c^{2}≤\frac{1}{54}\).

Log in to reply

See my note

Log in to reply

A barrel contains \(2n\) balls, numbered \(1\) to \(2n\). Choose three balls at random, one after the other, and with the balls replaced after each draw.

What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, is repeated?

Log in to reply

1/2

Log in to reply

Correct. Can you show your steps?

Log in to reply

Write the sum \(S_n= \displaystyle \sum_{k=0}^n{\frac{(-1)^{k}{n \choose{k}}}{k^{3}+ 9k^{2}+26k+24}}\) in the form \(\frac{p(n)}{q(n)}\) where \(p(n),q(n)\) are polynomials with integer coefficients.

Log in to reply

We first note that \(k^{3}+9k^{2}+26k+24=(k+2)(k+3)(k+4)\).Hence, \[S_n=\displaystyle \sum_{k=0}^n{\frac{(-1)^{k}{n \choose{k}}}{(k+2)(k+3)(k+4)}}\] \[=\displaystyle \sum_{k=0}^n{\frac{(-1)^{k}n!}{(n-k)k!(k+2)(k+3)(k+4)}}\] \[=\displaystyle \sum_{k=0}^n{\frac{(-1)^{k}(n+4)!}{(n-k!)(k+4!)}}\frac{(k+1)}{(n+1)(n+2)(n+3)(n+4)}\]

Let \(T_n=S_n(n+1)(n+2)(n+3)(n+4)=\small \displaystyle \sum_{k=0}^n{(-1)^{k}{n+4 \choose{k+4}}(k+1)}\).

We note that \(\small \displaystyle \sum_{k=0}^n{(-1)^{k}{n \choose{k}}}\).Now,

\(\displaystyle \sum_{k=0}^n{(-1)^{k}{n \choose{k}}k}\)\(=\displaystyle \sum_{k=1}^n{(-1)^{k}n{n-1 \choose{k-1}}+0}\)\(=-n \displaystyle \sum_{j=0}^{n-1}{(-1)^{j}{n-1 \choose{j}}=0}\).

\[T_n=\displaystyle \sum_{k=0}^n{(-1)^{k}{n+4 \choose{k+4}}(k+1)}\]. \[=\displaystyle \sum_{k=0}^n{(-1)^{k+4}{n+4 \choose{k+4}}(k+1)}\]. \[=\displaystyle \sum_{k=-4}^n{(-1)^{k+4}{n+4 \choose{k+4}}(k+1)-((-3)+2(n+4)-{n+4 \choose{2}}})\].

Substituting \(j=k+4\);

\[T_n=\displaystyle \sum_{j=0}^{n+4}{(-1)^{j}{n+4 \choose{j}}(j-3)-((-3)+2(n+4)-{n+4 \choose{2}}})\].

\(T_n=\displaystyle \sum_{j=0}^{n+4}{(-1)^{j}{n+4 \choose{j}}}-3 \displaystyle \sum_{j=0}^{n+4}{(-1)^{j}{n+4 \choose{j}}(j-3)-((-3)+2(n+4)-{n+4 \choose{2}}})\).

The first two terms are zero, hence

\(T_n={n+4 \choose{j}}-2n-8+3=\frac{(n+4)(n+3)}{2}-2n-5\).

Hence, \(T_n=\frac{(n+1)(n+2)}{2}\) and \(S_n=\frac{1}{2(n+3)(n+4)}\).

Phew!!

Log in to reply

Nicely done. There are a few mistakes in the solution but one may understand the correct part as he reads it.

Log in to reply

For any integer \(a\), prove that \(a^{37} \equiv a \pmod{1729}\).

Log in to reply

Its easy just apply keep applying FLT for various factors of 1729

Log in to reply

Prove that the function \(f(n)=\lfloor n+\sqrt{n}+\frac{1}{2} \rfloor\) where \(n\) is a positive integer, contains all numbers except perfect squares.

\((\lfloor \rfloor)\) represents the greatest integer function.

Log in to reply

There should be no perfect squares in the domain or the range?

Log in to reply

Range.

Log in to reply

Is n a natural number or something like that ? Because if n is any real then put n=0.5 . You will get f(0.5)=1 which is a perfect square.

Log in to reply

yes n is a natural number. Sorry forgot to mention it

Log in to reply

I have been trying this question but I have got no hint of even how to start. Could u give some starting line ?

Log in to reply

All the best to everyone writing RMO!

Log in to reply

All the best everyone! :)

Log in to reply

All the best everyone.!!

Log in to reply

All the best everyone ! :)

Log in to reply

@Nihar Mahajan @Saarthak Marathe @Svatejas Shivakumar @Harsh Shrivastava @Adarsh Kumar @Dev Sharma have your respective region's RMO results been declared ? Mumbai RMO results are declared and I am selected for INMO which is on January 17. Less than a month !!!! I think we should get back to work again. Lets start an INMO practise board just like this one as it helped me a lot in my RMO preparation. Hoping for a quick response

Log in to reply

BTW how many did you solve?

Log in to reply

Congo!

Btw my region results are not out yet.

Log in to reply

Thanks. Which is your region ? And what is the approx result date for the past few years ?

Log in to reply

Log in to reply

Congo dude! I am a bit busy right now!Sorry I can't take part in the discussion now!

Log in to reply

Congratulations!! The results have not been declared from my region (I don't think that I have much chances of qualifying this time. I did many silly mistakes :(

Log in to reply

I didn't qualify :(

Log in to reply

Hey !! That doesn't stop you from taking part in the discussions on INMO. Its okay ... Ur just in 8th standard, I came to know about RMO in 10th ...so u are at an advantage. And I am quite sure u will make it to the IMO by your 11-12th standard. I just don't know how u r able to solve olympiad questions being in 8th standard. So now bro .... u r targeting for IMO and thus u r participating in the INMO board ...

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Could someone please post some nice RMO-INMO level Geometry problems ? I have nothing to work on. Please not from the previous years.

Log in to reply

See the problems of this set. Can you suggest me some good books for Geometry? My geometry is horrible.

Log in to reply

Thanks for that set ! For geometry I have been solving only Challenges and Thrills ... I don't have something exclusively for geometry....

Log in to reply

@Harsh Shrivastava @Dev Sharma @Adarsh Kumar @Svatejas Shivakumar @Saarthak Marathe@Nihar Mahajan Where are u guys ?????? Please be active on the INMO practise board as well like u were on this board.... we need a lot of questions and a lot of solutions .....

Log in to reply

In the first question (sum(ω))^2>3sum(ω1*ω2) fails to hold? why?

Log in to reply