RMO Practice Board!

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,fa, b, c, d, e, f be real numbers such that the polynomial equation

x84x7+7x6+ax5+bx4+cx3+dx2+ex+f=0x^{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 ff

Note by Harsh Shrivastava
4 years, 6 months ago

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1 vote

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Comments

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Top Newest

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

Mayank Jha - 3 years, 7 months ago

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@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 .....

Shrihari B - 4 years, 4 months ago

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

Shrihari B - 4 years, 5 months ago

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

A Former Brilliant Member - 4 years, 4 months ago

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

Shrihari B - 4 years, 4 months ago

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@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

Shrihari B - 4 years, 5 months ago

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I didn't qualify :(

A Former Brilliant Member - 4 years, 4 months ago

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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 ...

Shrihari B - 4 years, 4 months ago

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@Shrihari B Thanks for your encouragement. I will definitely participate in the INMO board and help you reach IMO this year!

A Former Brilliant Member - 4 years, 4 months ago

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@A Former Brilliant Member Frankly speaking even INMO is impossible for me .... but I am just enjoying math. I have got an excuse of not studying for IIT-JEE till Jan 17(INMO).

Shrihari B - 4 years, 4 months ago

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@Shrihari B You're not an IIT-JEE aspirant?

A Former Brilliant Member - 4 years, 4 months ago

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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 :(

A Former Brilliant Member - 4 years, 5 months ago

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BTW how many did you solve?

Adarsh Kumar - 4 years, 5 months ago

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

Adarsh Kumar - 4 years, 5 months ago

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Congo!

Btw my region results are not out yet.

Harsh Shrivastava - 4 years, 5 months ago

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

Shrihari B - 4 years, 5 months ago

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@Shrihari B Chhattisgrah State.

Harsh Shrivastava - 4 years, 5 months ago

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All the best to everyone writing RMO!

A Former Brilliant Member - 4 years, 5 months ago

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All the best everyone ! :)

Shrihari B - 4 years, 5 months ago

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All the best everyone.!!

Saarthak Marathe - 4 years, 5 months ago

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All the best everyone! :)

Nihar Mahajan - 4 years, 5 months ago

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Let x,yx,y be real numbers such that x,y>1x,y > 1

Prove that x2y1+y2x18\dfrac{x^{2}}{y-1} + \dfrac{y^{2}}{ x-1} \geq 8

Harsh Shrivastava - 4 years, 5 months ago

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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

Shrihari B - 4 years, 5 months ago

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

Then it will be easy manipulate the expression.

Harsh Shrivastava - 4 years, 5 months ago

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Nice idea!

Nihar Mahajan - 4 years, 5 months ago

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

x2y1+y2x1y2y1+y2x14+y2x1    x2y14\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 x2y1\frac{x^{2}}{y-1} is 44 and it occurs if and only if x=y=2x=y=2.

Therefore, x2y1+y2x18\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 8.

A Former Brilliant Member - 4 years, 5 months ago

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Hmm you have not proven that min value of x2y1\frac{x^{2}}{y-1} is 4.

Harsh Shrivastava - 4 years, 5 months ago

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@Harsh Shrivastava Do you mean that I haven't proven that minimum value of y2y1\frac{y^{2}}{y-1} is 44?

A Former Brilliant Member - 4 years, 5 months ago

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Prove that the function f(n)=n+n+12f(n)=\lfloor n+\sqrt{n}+\frac{1}{2} \rfloor where nn is a positive integer, contains all numbers except perfect squares.

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

A Former Brilliant Member - 4 years, 5 months ago

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

Shrihari B - 4 years, 5 months ago

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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.

Shrihari B - 4 years, 5 months ago

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

A Former Brilliant Member - 4 years, 5 months ago

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

Saarthak Marathe - 4 years, 5 months ago

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Range.

A Former Brilliant Member - 4 years, 5 months ago

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For any integer aa, prove that a37a(mod1729)a^{37} \equiv a \pmod{1729}.

A Former Brilliant Member - 4 years, 5 months ago

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

Shrihari B - 4 years, 5 months ago

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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 :)

Shrihari B - 4 years, 5 months ago

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Done :)

A Former Brilliant Member - 4 years, 5 months ago

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Thanks

Shrihari B - 4 years, 5 months ago

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1991k(199019911992+199219911990)1991^{k}|(1990^{1991^{1992}}+1992^{1991^{1990}})

max(k)=?max(k)=?


Details:

bab|a\rightarrow aa is divisible by bb.

Akshat Sharda - 4 years, 5 months ago

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1991

Saarthak Marathe - 4 years, 5 months ago

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Yes..... Can you post solution ?

Akshat Sharda - 4 years, 5 months ago

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

Shrihari B - 4 years, 5 months ago

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@Shrihari B Shrihari,I have posted the general result of that combinatorics problem in its 'discuss solution' part. Now try to prove the general result .

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

Saarthak Marathe - 4 years, 5 months ago

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@Saarthak Marathe Hey Saarthak I haven't heard of Catalan numbers before. You need to elaborate a bit Could you give a proof for the number of ways ? It will help me understand better. Thanks in advance

Shrihari B - 4 years, 5 months ago

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Try this set guys.

Akshat Sharda - 4 years, 5 months ago

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

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

Harsh Shrivastava - 4 years, 5 months ago

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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 :)

Shrihari B - 4 years, 5 months ago

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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 BAC=90\angle BAC = 90^\circ , do you see why?

Nihar Mahajan - 4 years, 5 months ago

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@Nihar Mahajan FD=FC=FA . Great !

Shrihari B - 4 years, 5 months ago

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@Shrihari B Do u have some problems for practice ? I need some number theory. I am not that good or in other words I am bad in it

Shrihari B - 4 years, 5 months ago

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@Shrihari B I also need some NT probs...

Harsh Shrivastava - 4 years, 5 months ago

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@Harsh Shrivastava Even I need NT problems...

Nihar Mahajan - 4 years, 5 months ago

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@Nihar Mahajan Are you going to appear for RMO from mumbai region or from the pune region ?

Shrihari B - 4 years, 5 months ago

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@Shrihari B Pune region.

Nihar Mahajan - 4 years, 5 months ago

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Yup I also got 90 degrees.

Harsh Shrivastava - 4 years, 5 months ago

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Shrihari,we get that CF=2DFCF=2DF

Saarthak Marathe - 4 years, 5 months ago

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@Saarthak Marathe I don't think so check again.

Shrihari B - 4 years, 5 months ago

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@Shrihari B Applying Menelaus' theorem, ( (BE/EC)(CF/FD)(DA/AB)=1

Saarthak Marathe - 4 years, 5 months ago

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@Saarthak Marathe BE/EC is 2 and DA/AB is 1/2. So CF/FD is 1

Shrihari B - 4 years, 5 months ago

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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.

Shrihari B - 4 years, 5 months ago

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Its easy , note that BP=QC=sbBP=QC=s-b where ss is the semiperimeter of the triangle and bb is length of side opposite to B \angle B respectively. Since MM is mid point of BCBC , the result follows.

Nihar Mahajan - 4 years, 5 months ago

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Oh nice !

Shrihari B - 4 years, 5 months ago

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

A Former Brilliant Member - 4 years, 5 months ago

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@A Former Brilliant Member In the instructions given in question paper , it is said that protractors are not allowed.

Nihar Mahajan - 4 years, 5 months ago

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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.

Rakhi Bhattacharyya - 4 years, 5 months ago

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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

Shrihari B - 4 years, 5 months ago

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Find all pairs (m,n)(m,n) such that 2m+3n2^m+3^n is a perfect square. (A popular NT problem)

Nihar Mahajan - 4 years, 5 months ago

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Its easy.Only one solution (2,2)

Saarthak Marathe - 4 years, 5 months ago

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Wrong. Check again.

Nihar Mahajan - 4 years, 5 months ago

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@Nihar Mahajan I think he means (4,2).

A Former Brilliant Member - 4 years, 5 months ago

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@A Former Brilliant Member Ya I meant (4,2). Sorry,typing error. :P

Saarthak Marathe - 4 years, 5 months ago

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@Saarthak Marathe Its okay , please post your method. :)

Nihar Mahajan - 4 years, 5 months ago

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@Nihar Mahajan Taking mod 4 we get that m,n should be even. Then using gcd(2m,3n)=1 gcd({2}^{m},3^{n})=1 , we use the general formula for pythagorean triplets. Taking various cases,we get that only m=4,n=2 m=4,n=2 are the solutions.

Saarthak Marathe - 4 years, 5 months ago

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If mm is a positive real number and satisfies the equation 2m=1+1m+m2142m=1+\frac{1}{m}+\sqrt{m^{2}-\frac{1}{4}} Find mm.

Shivam Jadhav - 4 years, 5 months ago

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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

Shrihari B - 4 years, 5 months ago

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Let p>3p>3 be a prime number. Suppose k=1p11k=ab\displaystyle \sum_{k=1}^{p-1}{\frac{1}{k}}=\frac{a}{b} where gcd(a,b)=1\gcd(a,b)=1. Prove that aa is divisible by p2p^{2}.

A Former Brilliant Member - 4 years, 5 months ago

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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 ?

Shrihari B - 4 years, 5 months ago

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Please post the proof, thanks!

Harsh Shrivastava - 4 years, 5 months ago

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@Harsh Shrivastava My proof is incomplete I could not prove that a is divisible by p^2

Shrihari B - 4 years, 5 months ago

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@Shrihari B Post the proof that a is divisible by p, Thanks!

Harsh Shrivastava - 4 years, 5 months ago

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@Harsh Shrivastava Take LCM on the LHS and try proving that each term in the numerator leaves a different residue modulo p. Do it by contradiction. Once you are able to prove that, the rest is clear. If it is difficult, first take a smaller example say p=7. Then work it out

Shrihari B - 4 years, 5 months ago

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I have the solution only till aa is divisible by pp. To prove that aa is divisible by p2p^{2} is given below the solution as a challenge.

A Former Brilliant Member - 4 years, 5 months ago

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Write the sum Sn=k=0n(1)k(nk)k3+9k2+26k+24S_n= \displaystyle \sum_{k=0}^n{\frac{(-1)^{k}{n \choose{k}}}{k^{3}+ 9k^{2}+26k+24}} in the form p(n)q(n)\frac{p(n)}{q(n)} where p(n),q(n)p(n),q(n) are polynomials with integer coefficients.

A Former Brilliant Member - 4 years, 5 months ago

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We first note that k3+9k2+26k+24=(k+2)(k+3)(k+4)k^{3}+9k^{2}+26k+24=(k+2)(k+3)(k+4).Hence, Sn=k=0n(1)k(nk)(k+2)(k+3)(k+4)S_n=\displaystyle \sum_{k=0}^n{\frac{(-1)^{k}{n \choose{k}}}{(k+2)(k+3)(k+4)}} =k=0n(1)kn!(nk)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)}} =k=0n(1)k(n+4)!(nk!)(k+4!)(k+1)(n+1)(n+2)(n+3)(n+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 Tn=Sn(n+1)(n+2)(n+3)(n+4)=k=0n(1)k(n+4k+4)(k+1)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 k=0n(1)k(nk)\small \displaystyle \sum_{k=0}^n{(-1)^{k}{n \choose{k}}}.Now,

k=0n(1)k(nk)k\displaystyle \sum_{k=0}^n{(-1)^{k}{n \choose{k}}k}=k=1n(1)kn(n1k1)+0=\displaystyle \sum_{k=1}^n{(-1)^{k}n{n-1 \choose{k-1}}+0}=nj=0n1(1)j(n1j)=0=-n \displaystyle \sum_{j=0}^{n-1}{(-1)^{j}{n-1 \choose{j}}=0}.

Tn=k=0n(1)k(n+4k+4)(k+1)T_n=\displaystyle \sum_{k=0}^n{(-1)^{k}{n+4 \choose{k+4}}(k+1)}. =k=0n(1)k+4(n+4k+4)(k+1)=\displaystyle \sum_{k=0}^n{(-1)^{k+4}{n+4 \choose{k+4}}(k+1)}. =k=4n(1)k+4(n+4k+4)(k+1)((3)+2(n+4)(n+42))=\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+4j=k+4;

Tn=j=0n+4(1)j(n+4j)(j3)((3)+2(n+4)(n+42))T_n=\displaystyle \sum_{j=0}^{n+4}{(-1)^{j}{n+4 \choose{j}}(j-3)-((-3)+2(n+4)-{n+4 \choose{2}}}).

Tn=j=0n+4(1)j(n+4j)3j=0n+4(1)j(n+4j)(j3)((3)+2(n+4)(n+42))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

Tn=(n+4j)2n8+3=(n+4)(n+3)22n5T_n={n+4 \choose{j}}-2n-8+3=\frac{(n+4)(n+3)}{2}-2n-5.

Hence, Tn=(n+1)(n+2)2T_n=\frac{(n+1)(n+2)}{2} and Sn=12(n+3)(n+4)S_n=\frac{1}{2(n+3)(n+4)}.

Phew!!

A Former Brilliant Member - 4 years, 5 months ago

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

Saarthak Marathe - 4 years, 5 months ago

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A barrel contains 2n2n balls, numbered 11 to 2n2n. 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?

A Former Brilliant Member - 4 years, 5 months ago

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1/2

Saarthak Marathe - 4 years, 5 months ago

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Correct. Can you show your steps?

A Former Brilliant Member - 4 years, 5 months ago

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Prove that 2<(n+2n+1)(n+1) 2 < (\dfrac{n+2}{n+1}) ^ {(n+1)} for all integers n>0n > 0.

Harsh Shrivastava - 4 years, 5 months ago

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

Shrihari B - 4 years, 5 months ago

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Prove that

1<11001+11002+.........+13001<431<\frac { 1 }{ 1001 } +\frac { 1 }{ 1002 } +.........+\frac { 1 }{ 3001 } <\frac { 4 }{ 3 }

Satyajit Ghosh - 4 years, 5 months ago

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

Saarthak Marathe - 4 years, 5 months ago

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

Satyajit Ghosh - 4 years, 5 months ago

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@Satyajit Ghosh Can u tell me how to add an image in the answer?

Saarthak Marathe - 4 years, 5 months ago

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@Saarthak Marathe 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 is Alternate text 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 This is an image file :D. Here's how it appears:

this is a copy paste from this note on moderators

Satyajit Ghosh - 4 years, 5 months ago

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@Satyajit Ghosh I think we can use calculus in rmo .But if we use it we need to be very careful in our steps and not miss any points.

Saarthak Marathe - 4 years, 5 months ago

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If a,b0a,b \ge 0. Prove that 2(a(a+b)3+ba2+b2)3(a2+b2)\sqrt{2}(\sqrt{a(a+b)^{3}}+b \sqrt{a^{2}+b^{2}}) \le 3(a^{2}+b^{2}).

A Former Brilliant Member - 4 years, 5 months ago

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Show that n5+n4+1n^{5}+n^{4}+1 is not prime for n>1n>1.

A Former Brilliant Member - 4 years, 5 months ago

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n2+n+1n^{2} + n + 1 is factor of the given expression.Hence the result is obvious.

Harsh Shrivastava - 4 years, 5 months ago

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Let x,yx,y and zz be positive real numbers such that x+y+z=1x+y+z=1. Prove that yz1+x+zx1+y+xy1+z14\frac{yz}{1+x}+\frac{zx}{1+y}+\frac{xy}{1+z} \le \frac{1}{4}.

A Former Brilliant Member - 4 years, 5 months ago

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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.

Saarthak Marathe - 4 years, 5 months ago

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If a,b,c,d,ea,b,c,d,e are real numbers, prove that the roots of x5+ax4+bx3+cx2+dx+e=0x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e=0 cannot all be real if 2a2<5b2a^{2}<5b.

A Former Brilliant Member - 4 years, 5 months ago

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Let roots be p,q,r,s,tp,q,r,s,t.

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

Therefore Sum of squares of roots = a22ba^{2} - 2b

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

a22ba25a^{2} -2b \geq \frac{a^{2}}{5}

    2a25b\implies 2a^{2} \geq 5b

Therefore the polynomial cannot have all roots real if 5b>2a2 5b > 2a^{2}.

Harsh Shrivastava - 4 years, 5 months ago

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Let a,bR,a0a,b \in R, a \neq 0. Show that a2+b2+1a2+ba3a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{b}{a} \ge \sqrt{3}.

A Former Brilliant Member - 4 years, 5 months ago

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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.

Shrihari B - 4 years, 5 months ago

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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.

Saarthak Marathe - 4 years, 5 months ago

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

Shrihari B - 4 years, 5 months ago

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I didn't get it.

Saarthak Marathe - 4 years, 5 months ago

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An easy INMO geometry problem:

Let ABCABC be a triangle with circumcircle Γ\Gamma. Let MM be a point in the interior of triangle ABC ABC which is also on the bisector of A\angle A. Let AM,BM,CMAM, BM, CM meet Γ \Gamma in A1,B1,C1 A_{1}, B_{1}, C_{1} respectively. Suppose PP is the point of intersection of A1C1 A_{1}C_{1} with ABAB and QQ is the point of intersection of A1B1A_{1}B_{1} with ACAC. Prove that PQPQ is parallel to BCBC .

Nihar Mahajan - 4 years, 5 months ago

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

Prove that ab+bc+ca34ab+bc+ca \leq \frac{3}{4}

Harsh Shrivastava - 4 years, 5 months ago

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Using am gm

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

a + b + c > 3/2

now using cauchy,

3(a2+b2+c2)>(a+b+c)23(a^2 + b^2 + c^2) > (a + b + c)^2

so

a2+b2+c2>3/4a^2 + b^2 + c^2 > 3/4

using cauchy again,

(a2+b2+c2)2>(ab+bc+ca)2(a^2 + b^2 + c^2)^2 > (ab + bc + ca)^2

3/4 > ab + bc + ca

Dev Sharma - 4 years, 5 months ago

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

Saarthak Marathe - 4 years, 5 months ago

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You got the following two inequalities,a2+b2+c2>ab+bc+caanda2+b2+c2>33a^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,34>ab+bc+ca\dfrac{3}{4}>ab+bc+ca?

Adarsh Kumar - 4 years, 5 months ago

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@Adarsh Kumar using cauchy again, like this

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

sum is cyclic

Dev Sharma - 4 years, 5 months ago

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Nice solution. You could also have used the fact that (a+b+c)23(ab+bc+ca)(a+b+c)^{2} \ge 3(ab+bc+ca)

A Former Brilliant Member - 4 years, 5 months ago

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

Harsh Shrivastava - 4 years, 5 months ago

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Harsh,could you please add a figure?

Adarsh Kumar - 4 years, 5 months ago

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Hint: Let QPCO={R}\overrightarrow{QP} \cap \overleftrightarrow{CO} = \{R\} and COΠ={S}\overleftrightarrow{CO} \cap \Pi = \{S\} and then it will suffice to prove that ΔCRPΔCAS\Delta CRP \sim \Delta CAS

Nihar Mahajan - 4 years, 5 months ago

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Find the least positive integer n such that 2549 divides n25452541n^{2545} - 2541

Dev Sharma - 4 years, 5 months ago

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Please post the solution , Thanks!

Harsh Shrivastava - 4 years, 5 months ago

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Let ABCABC be a triangle and PP be a point inside it.Let x,y,zx,y,z denote the lengths of PAPA,PBPB,PCPC.Let a,b,ca,b,c be the lengths of sides BCBC,CACA,ABAB.Find all points PP such that axy+byz+czx=abcaxy+byz+czx=abc.

Souryajit Roy - 4 years, 5 months ago

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You have a set of consecutive positive integers,{1,2,3,...,199,200}\{1,2,3,...,199,200\},if you choose 101101 numbers from these at random,then prove that there at-least two numbers of which one divides the other.

Adarsh Kumar - 4 years, 5 months ago

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Find all positive integer solutions aa, bb and cc such that

7a2+11b2=c77.\dfrac {7}{a^2} + \dfrac {11}{b^2} = \dfrac {c}{77}.

Sharky Kesa - 4 years, 5 months ago

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

7a2+11b2=c77.\dfrac {7 }{a^2} + \dfrac {11 }{b^2} = \dfrac{c}{77}.

Case 1: c>1c>1

This forces a2<7 and b2<11a^{2} < 7 \text{ and } b^{2} < 11

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

Case 2: c77c \leq 77

Since c77c \leq 77 this implies c771\dfrac{c}{77} \leq 1.

Let c77=1k\dfrac{c}{77} = \dfrac{1}{k} for some integer k1k \geq 1.

    c=77k\implies c = \dfrac{77}{k}

Now since cc is a positive integer , this forces k | 77k \text{ | } 77.

Possible values for k=1,7,11,77k = 1,7,11,77.

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

Harsh Shrivastava - 4 years, 5 months ago

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Let a,b,c,dNa,b,c,d \in \mathbb{N} such that abcda \ge b \ge c \ge d and d0d \neq 0. Show that the equation x4ax3bx2cxd=0x^4 - ax^3 - bx^2 - cx -d = 0 has no integer solution in xx.

Please post a "complete solution" like you would do in RMO.

Nihar Mahajan - 4 years, 5 months ago

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let n be an integral solution (if possible). then n [n^3-an^2-bn-c] = d. Now for the equation to hold n divides d. so n is less than or equal to d. If n is negative -ax^3 is positive and is greater than bx^2 as a is greater than b.similarly for the next term and the total value becomes positive but known the value is 0.hence our assumption that n is negative is wrong .so n is positive and less than d. Now as n<d <a, n^4<an^3 and -bn^2 and -cx and -d all are negative,so the sum is also negative which was again 0. hence contradiction. So no integral roots exists.

badisa books - 1 month, 1 week ago

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Isn't this from INMO 2013 ??

Shrihari B - 4 years, 4 months ago

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Yes it is.

Nihar Mahajan - 4 years, 4 months ago

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@Nihar Mahajan Hey why aren't u participating in the INMO practice board ?

Shrihari B - 4 years, 4 months ago

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I'll have a crack at it. As d0d \neq 0, x0x \neq 0.

Denote y=1xy = \frac{1}{x}.

We have

x4=ax3+bx2+cx+dx^4 = ax^3 + bx^2 + cx + d \quad \quad

1=ay+by2+cy3+dy4(1)1 = ay + by^2 + cy^3 + dy^4 \quad \quad (1)

Now, we separate this into 2 cases:

Case 1:

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

an+bn2+cn3+dn4=1\dfrac {-a}{n} + \dfrac {b}{n^2} + \dfrac {-c}{n^3} + \dfrac {d}{n^4} = 1

an+bn2+cn+dn4=1\dfrac {-an+b}{n^2}+\dfrac{-cn+d}{n^4}=1

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

Case 2:

x=nx = n where nNn \in \mathbb{N}, then y=1ny=\dfrac{1}{n}. Substituting this into (1)(1), we get

an+bn2+cn3+dn4=1\dfrac {a}{n} + \dfrac {b}{n^2} + \dfrac {c}{n^3} + \dfrac {d}{n^4} = 1

This implies that n1n \neq 1, and

an<1\dfrac {a}{n} < 1

1an+an2+an3+an4<i=1ani=an11 \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

na>1\dfrac {n}{a} > 1

n1a<1\dfrac {n-1}{a} < 1

There are no such nn that satisfy the above two inequalities. Thus, no solutions exist.

Sharky Kesa - 4 years, 5 months ago

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If a circle goes through point AA of a parallelogram ABCDABCD, cuts the two sides AB,ADAB,AD and the diagonal ACAC at points P,RP,R and QQ respectively. Prove that (AP×AB)+(AR×AD)=AQ×AC(AP×AB)+(AR×AD)=AQ×AC.

A Former Brilliant Member - 4 years, 5 months ago

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

Let ASAS be the diameter of given circle.

AB+AD=ACAB+AD=AC

Therefore taking dot product withASAS

AB.AS+AD.AS=AC.ASAB.AS+AD.AS=AC.AS

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

ABAP+ARAD=AQAC|AB|*|AP|+|AR|*|AD|=|AQ|*|AC|

Saarthak Marathe - 4 years, 5 months ago

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Let pp and qq be distinct primes. Show that

j=1(p1)/2[qjp]+j=1(q1)/2[pjq]=(p1)(q1)4\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.

A Former Brilliant Member - 4 years, 5 months ago

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Let S=(x,y)x,yN,1x(p1)/2,1y(q1)/2S={(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 SS can be partitioned into two sets S1S_1 and S2S_2 according as qx>pyqx>py or qx<pyqx<py. Note that there are no pairs in such SS such that qx=pyqx=py.

The set S1S_1 can be described as the set of all pairs (x,y)(x,y) satisfying 1x(p1)/2,1y<qx/p1 \le x \le (p-1)/2, \quad 1 \le y < qx/p. Then S1S_1 has j=1(p1)/2qjp\displaystyle \sum_{j=1}^{(p-1)/2}{\frac{qj}{p}} elements.

Similarly, S2S_2 has i=1(q1)/2piq\displaystyle \sum_{i=1}^{(q-1)/2}{\frac{pi}{q}}. Hence, we get the required result.

A Former Brilliant Member - 4 years, 5 months ago

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Given seven arbitrary distinct real numbers, show that there exist two numbers xx and yy such thst 0<xy1+xy<130<\dfrac{x-y}{1+xy}<\dfrac{1}{\sqrt{3}}.

A Former Brilliant Member - 4 years, 5 months ago

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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).

Shrihari B - 4 years, 5 months ago

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Find all primes x,yx,y such that xyyx=xy219x^{y}-y^{x}=xy^{2}-19.Enjoy solving this.

Souryajit Roy - 4 years, 5 months ago

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Using FLT, we have

xyx(mody)x19(mody)x+190(mody)x^y \equiv x \pmod{y} \Rightarrow x \equiv -19 \pmod {y} \Rightarrow x+19 \equiv 0 \pmod{y}

yxy(modx)y19(modx)19y0(modx)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=2x=2 or y=2y=2. We now separate this into 2 cases:

Case 1: y=2y=2

When y=2y=2, we have 192=170(modx)19 - 2 = 17 \equiv 0 \pmod{x}, which implies x=17x=17. The solution in this case is (17,2)(17, 2). Checking, however, we find that this clearly doesn't work. So, there are no solutions in this case.

Case 2: x=2x=2

When x=2x=2, we have 19+2=210(mody)19+2 = 21 \equiv 0 \pmod {y}, which implies y=3y=3 or y=7y=7. Checking, we find both these solutions work. Thus, 2 solutions exist in this case: (2,3)(2, 3) and (2,7)(2,7).

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

Sharky Kesa - 4 years, 5 months ago

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