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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 \(a, b, c, d, e, f\) 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 \(f\)

Note by Harsh Shrivastava
10 months, 1 week ago

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Since 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? Adarsh Kumar · 10 months, 1 week ago

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@Adarsh Kumar In your solution you need prove that \[(\sum_{cyclic}ω)^{2}-\sum_{cyclic}ω_{1}ω_{2}=\sum_{cyclic}ω^{2}\] Shivam Jadhav · 10 months, 1 week ago

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@Shivam Jadhav That's an identity that is true for all numbers, real and imaginary. And there should be a 2 infront of that second summation. Trevor Arashiro · 9 months ago

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@Trevor Arashiro Yes , its one of the short Newton identities. Nihar Mahajan · 9 months ago

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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 . Anish Harsha · 10 months ago

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@Anish Harsha 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 Souryajit Roy · 10 months ago

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

\[\dfrac {7}{a^2} + \dfrac {11}{b^2} = \dfrac {c}{77}.\] Sharky Kesa · 10 months ago

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

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

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 \(a,b\) Harsh Shrivastava · 10 months ago

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If \(x\) and \(y\) are integers such that \(y^2+3(xy)^2=30x^2+517\), Prove that \(3(xy)^2=588\) Lakshya Sinha · 10 months, 1 week ago

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@Lakshya Sinha 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 \(3(xy)^{2}= 588\).

\(\QED \) Harsh Shrivastava · 10 months ago

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Let \(x,y\) and \(z\) be positive real numbers such that \(x+y+z=1\). Prove that \(\frac{yz}{1+x}+\frac{zx}{1+y}+\frac{xy}{1+z} \le \frac{1}{4}\). Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 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 · 10 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 · 10 months ago

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@Saarthak Marathe Can u send the solution of this question ? I have been trying. No progress .... Shrihari B · 9 months, 4 weeks ago

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@Shrihari B I didn't get it. Saarthak Marathe · 9 months, 3 weeks ago

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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. Adarsh Kumar · 10 months ago

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Find all primes \(p\) such that for all prime \(q\), the remainder of \(p\) upon divison by \(q\) is squarefree. Dev Sharma · 10 months, 1 week ago

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@Dev Sharma I think that 3 is the only prime number. I used the fact that all primes>3 is of the form \(\pm 1 \pmod{6}\). Svatejas Shivakumar · 9 months, 4 weeks ago

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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 \(\angle BAC\). Harsh Shrivastava · 9 months, 4 weeks ago

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@Harsh Shrivastava 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 · 9 months, 4 weeks ago

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@Nihar Mahajan Any solution is welcome. How did u solve using Trigonometry ? Shrihari B · 9 months, 3 weeks ago

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@Shrihari B Let \(\angle FDA = \angle FAD = \alpha \ , \ \angle AEC=\gamma\).Let \(FD=FA=x \ , \ CE=y \ , \ BE=2y\). Using Sine Rule in \(\Delta ADF\) , we get \(AD=2x\cos(\alpha)\) and thus \(AB=4x\cos(\alpha)\). By angle chase we have \(\angle CFE= 180-2\alpha\) and again applying Sine Rule to \(\Delta CEF\) , we get \(FC=\dfrac{y\sin(\gamma)}{sin(2\alpha)}\). Now apply Sine Rule to \(\Delta ABE\) and knowing that \(\sin(180^\circ - \gamma)= \sin(\gamma)\) , we get \(x=\dfrac{y\sin(\gamma)}{sin(2\alpha)}= FC\) . This means \(F\) is midpoint of \(DC\) , and we get the desired result. Nihar Mahajan · 9 months, 3 weeks ago

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@Nihar Mahajan Nice work there ! Shrihari B · 9 months, 3 weeks ago

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@Shrihari B 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? Nihar Mahajan · 9 months, 4 weeks ago

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@Nihar Mahajan FD=FC=FA . Great ! Shrihari B · 9 months, 4 weeks 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 · 9 months, 4 weeks ago

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@Shrihari B I also need some NT probs... Harsh Shrivastava · 9 months, 4 weeks ago

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@Harsh Shrivastava Even I need NT problems... Nihar Mahajan · 9 months, 4 weeks ago

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@Nihar Mahajan Are you going to appear for RMO from mumbai region or from the pune region ? Shrihari B · 9 months, 4 weeks ago

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@Shrihari B Pune region. Nihar Mahajan · 9 months, 4 weeks ago

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@Shrihari B Yup I also got 90 degrees. Harsh Shrivastava · 9 months, 4 weeks ago

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@Shrihari B Shrihari,we get that \(CF=2DF\) Saarthak Marathe · 9 months, 4 weeks ago

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@Saarthak Marathe I don't think so check again. Shrihari B · 9 months, 4 weeks ago

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@Shrihari B Applying Menelaus' theorem, ( (BE/EC)(CF/FD)(DA/AB)=1 Saarthak Marathe · 9 months, 4 weeks ago

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@Saarthak Marathe BE/EC is 2 and DA/AB is 1/2. So CF/FD is 1 Shrihari B · 9 months, 4 weeks ago

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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 \(p^{2}\). Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 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 · 9 months, 4 weeks ago

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@Shrihari B Please post the proof, thanks! Harsh Shrivastava · 9 months, 3 weeks ago

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@Harsh Shrivastava My proof is incomplete I could not prove that a is divisible by p^2 Shrihari B · 9 months, 3 weeks ago

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@Shrihari B Post the proof that a is divisible by p, Thanks! Harsh Shrivastava · 9 months, 3 weeks 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 · 9 months, 3 weeks ago

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@Shrihari B 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. Svatejas Shivakumar · 9 months, 4 weeks ago

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If \(a,b \ge 0\). Prove that \(\sqrt{2}(\sqrt{a(a+b)^{3}}+b \sqrt{a^{2}+b^{2}}) \le 3(a^{2}+b^{2})\). Svatejas Shivakumar · 10 months ago

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

Prove that \(ab+bc+ca \leq \frac{3}{4}\) Harsh Shrivastava · 10 months ago

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@Harsh Shrivastava 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 Dev Sharma · 10 months ago

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@Dev Sharma Nice solution. You could also have used the fact that \((a+b+c)^{2} \ge 3(ab+bc+ca)\) Svatejas Shivakumar · 10 months ago

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@Dev Sharma Dev,your solution is incorrect. Check the solution again,you will find your flaw. Saarthak Marathe · 10 months ago

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@Dev Sharma 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,\[\dfrac{3}{4}>ab+bc+ca\]? Adarsh Kumar · 10 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 · 10 months ago

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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. Nihar Mahajan · 10 months ago

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@Nihar Mahajan 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. Sharky Kesa · 10 months ago

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@Nihar Mahajan Isn't this from INMO 2013 ?? Shrihari B · 9 months ago

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@Shrihari B Yes it is. Nihar Mahajan · 9 months ago

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@Nihar Mahajan Hey why aren't u participating in the INMO practice board ? Shrihari B · 9 months ago

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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. Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 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. Svatejas Shivakumar · 9 months, 4 weeks ago

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Find all primes \(x,y\) such that \(x^{y}-y^{x}=xy^{2}-19\).Enjoy solving this. Souryajit Roy · 10 months ago

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@Souryajit Roy 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 \((2,7)\). Sharky Kesa · 10 months ago

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@Sharky Kesa would you please explain how x = -19 mody Dev Sharma · 10 months ago

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@Dev Sharma Putting \(x^y-y^x=xy^2-19\) in \(\pmod{y}\), we get \(x=-19 \pmod{y}\). Sharky Kesa · 10 months ago

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@Souryajit Roy 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). Adarsh Kumar · 10 months ago

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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 \(\dfrac{41}{42}\) Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 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. Svatejas Shivakumar · 10 months ago

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

\[x + y + z + \dfrac{3}{x-1}+ \dfrac{3}{y-1} + \dfrac{3}{z-1} = 2(\sqrt{x+2} + \sqrt{y+ 2} + \sqrt{z+2})\] Harsh Shrivastava · 10 months ago

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@Harsh Shrivastava https://brilliant.org/problems/three-variables-one-equation/?group=4bQcwX9pLM2R#!/solution-comments/106741/ Surya Prakash · 10 months ago

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

Prove that \(\dfrac{x^{2}}{y-1} + \dfrac{y^{2}}{ x-1} \geq 8\) Harsh Shrivastava · 9 months, 3 weeks ago

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@Harsh Shrivastava 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 · 9 months, 3 weeks ago

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

Then it will be easy manipulate the expression. Harsh Shrivastava · 9 months, 3 weeks ago

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@Shrihari B Nice idea! Nihar Mahajan · 9 months, 3 weeks ago

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@Harsh Shrivastava 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, \(\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 8\). Svatejas Shivakumar · 9 months, 3 weeks ago

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@Svatejas Shivakumar Hmm you have not proven that min value of \(\frac{x^{2}}{y-1}\) is 4. Harsh Shrivastava · 9 months, 3 weeks ago

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@Harsh Shrivastava Do you mean that I haven't proven that minimum value of \(\frac{y^{2}}{y-1}\) is \(4\)? Svatejas Shivakumar · 9 months, 3 weeks 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 · 9 months, 4 weeks ago

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@Shrihari B Done :) Svatejas Shivakumar · 9 months, 4 weeks ago

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@Svatejas Shivakumar Thanks Shrihari B · 9 months, 4 weeks ago

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

\(max(k)=?\)


Details:

\(b|a\rightarrow\) \(a\) is divisible by \(b\).

Akshat Sharda · 9 months, 4 weeks ago

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@Akshat Sharda 1991 Saarthak Marathe · 9 months, 3 weeks ago

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@Saarthak Marathe Yes..... Can you post solution ? Akshat Sharda · 9 months, 3 weeks ago

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@Saarthak Marathe How ? Could u give some rough outline of your solution ? Shrihari B · 9 months, 3 weeks 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 · 9 months, 3 weeks 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 · 9 months, 3 weeks ago

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Try this set guys. Akshat Sharda · 9 months, 4 weeks 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 · 10 months ago

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@Shrihari B 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. Nihar Mahajan · 9 months, 4 weeks ago

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@Nihar Mahajan Oh nice ! Shrihari B · 9 months, 4 weeks ago

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@Nihar Mahajan Are protractors allowed for RMO? Nothing about it is mentioned in the website and the hall ticket. Svatejas Shivakumar · 9 months, 4 weeks ago

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@Svatejas Shivakumar In the instructions given in question paper , it is said that protractors are not allowed. Nihar Mahajan · 9 months, 4 weeks 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 · 10 months ago

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@Rakhi Bhattacharyya 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 · 10 months ago

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Find all pairs \((m,n)\) such that \(2^m+3^n\) is a perfect square. (A popular NT problem) Nihar Mahajan · 10 months ago

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@Nihar Mahajan Its easy.Only one solution (2,2) Saarthak Marathe · 9 months, 3 weeks ago

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@Saarthak Marathe Wrong. Check again. Nihar Mahajan · 9 months, 3 weeks ago

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@Nihar Mahajan I think he means (4,2). Svatejas Shivakumar · 9 months, 3 weeks ago

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@Svatejas Shivakumar Ya I meant (4,2). Sorry,typing error. :P Saarthak Marathe · 9 months, 3 weeks ago

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@Saarthak Marathe Its okay , please post your method. :) Nihar Mahajan · 9 months, 3 weeks ago

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@Nihar Mahajan Taking mod 4 we get that m,n should be even. Then using \( gcd({2}^{m},3^{n})=1 \), we use the general formula for pythagorean triplets. Taking various cases,we get that only \( m=4,n=2\) are the solutions. Saarthak Marathe · 9 months, 3 weeks ago

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If \(m\) is a positive real number and satisfies the equation \[2m=1+\frac{1}{m}+\sqrt{m^{2}-\frac{1}{4}}\] Find \(m\). Shivam Jadhav · 10 months ago

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@Shivam Jadhav 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 · 9 months, 4 weeks ago

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Prove that \[ 2 < (\dfrac{n+2}{n+1}) ^ {(n+1)}\] for all integers \(n > 0\). Harsh Shrivastava · 10 months ago

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@Harsh Shrivastava Well its easy just use Bernoulli's inequality that (1+x)^n > 1+nx Shrihari B · 10 months ago

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Prove that

\(1<\frac { 1 }{ 1001 } +\frac { 1 }{ 1002 } +.........+\frac { 1 }{ 3001 } <\frac { 4 }{ 3 } \) Satyajit Ghosh · 10 months ago

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@Satyajit Ghosh It is easy. I got a calculus and a non calculus answer Saarthak Marathe · 10 months ago

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@Saarthak Marathe Since Rmo is a precalculus olympiad, answer without calculus will be better:P Satyajit Ghosh · 10 months ago

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@Satyajit Ghosh Can u tell me how to add an image in the answer? Saarthak Marathe · 10 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 · 10 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 · 10 months ago

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Show that \(n^{5}+n^{4}+1\) is not prime for \(n>1\). Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar \(n^{2} + n + 1\) is factor of the given expression.Hence the result is obvious. Harsh Shrivastava · 10 months ago

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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 \(2a^{2}<5b\). Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 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 \( 5b > 2a^{2}\). Harsh Shrivastava · 10 months ago

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Let \(a,b \in R, a \neq 0\). Show that \(a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{b}{a} \ge \sqrt{3}\). Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 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 · 10 months ago

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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 \(BC\) . Nihar Mahajan · 10 months ago

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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. Harsh Shrivastava · 10 months ago

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@Harsh Shrivastava Harsh,could you please add a figure? Adarsh Kumar · 10 months ago

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@Adarsh Kumar Hint: Let \(\overrightarrow{QP} \cap \overleftrightarrow{CO} = \{R\}\) and \(\overleftrightarrow{CO} \cap \Pi = \{S\}\) and then it will suffice to prove that \(\Delta CRP \sim \Delta CAS\) Nihar Mahajan · 10 months ago

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Find the least positive integer n such that 2549 divides \(n^{2545} - 2541\) Dev Sharma · 10 months ago

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@Dev Sharma Please post the solution , Thanks! Harsh Shrivastava · 9 months, 4 weeks ago

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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 \(axy+byz+czx=abc\). Souryajit Roy · 10 months ago

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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 \((AP×AB)+(AR×AD)=AQ×AC\). Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 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]

\(|AB|*|AP|+|AR|*|AD|=|AQ|*|AC| \) Saarthak Marathe · 9 months, 4 weeks ago

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Comment deleted 9 months ago

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@Saarthak Marathe You can also use Ptolemy's theorem to the quadrilateral \(PQRA\) and then replace the sides of \(\Delta PQR\) by the corresponding sides of the similar triangle \(CBA\).( \(\Delta PQR \sim \Delta CBA\) by \(AAA\) similarity.) Svatejas Shivakumar · 9 months, 4 weeks ago

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Given seven arbitrary distinct real numbers, show that there exist two numbers \(x\) and \(y\) such thst \(0<\dfrac{x-y}{1+xy}<\dfrac{1}{\sqrt{3}}\). Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 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 · 10 months ago

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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. Lakshya Sinha · 10 months, 1 week 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 · 9 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 · 9 months, 2 weeks ago

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@Shrihari B See the problems of this set. Can you suggest me some good books for Geometry? My geometry is horrible. Svatejas Shivakumar · 9 months ago

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@Svatejas Shivakumar Thanks for that set ! For geometry I have been solving only Challenges and Thrills ... I don't have something exclusively for geometry.... Shrihari B · 9 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 · 9 months, 2 weeks ago

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@Shrihari B BTW how many did you solve? Adarsh Kumar · 9 months, 2 weeks ago

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@Shrihari B I didn't qualify :( Svatejas Shivakumar · 9 months ago

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@Svatejas Shivakumar 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 · 9 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! Svatejas Shivakumar · 9 months ago

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@Svatejas Shivakumar 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 · 9 months ago

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@Shrihari B You're not an IIT-JEE aspirant? Svatejas Shivakumar · 9 months ago

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@Shrihari B 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 :( Svatejas Shivakumar · 9 months, 2 weeks ago

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@Shrihari B Congo dude! I am a bit busy right now!Sorry I can't take part in the discussion now! Adarsh Kumar · 9 months, 2 weeks ago

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@Shrihari B Congo!

Btw my region results are not out yet. Harsh Shrivastava · 9 months, 2 weeks ago

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@Harsh Shrivastava Thanks. Which is your region ? And what is the approx result date for the past few years ? Shrihari B · 9 months, 2 weeks ago

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@Shrihari B Chhattisgrah State. Harsh Shrivastava · 9 months, 2 weeks ago

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All the best to everyone writing RMO! Svatejas Shivakumar · 9 months, 3 weeks ago

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@Svatejas Shivakumar All the best everyone ! :) Shrihari B · 9 months, 3 weeks ago

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@Svatejas Shivakumar All the best everyone.!! Saarthak Marathe · 9 months, 3 weeks ago

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@Svatejas Shivakumar All the best everyone! :) Nihar Mahajan · 9 months, 3 weeks ago

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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. Svatejas Shivakumar · 9 months, 3 weeks ago

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@Svatejas Shivakumar 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 · 9 months, 2 weeks ago

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@Svatejas Shivakumar 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 · 9 months, 3 weeks ago

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@Shrihari B yes n is a natural number. Sorry forgot to mention it Svatejas Shivakumar · 9 months, 3 weeks ago

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@Svatejas Shivakumar There should be no perfect squares in the domain or the range? Saarthak Marathe · 9 months, 3 weeks ago

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@Saarthak Marathe Range. Svatejas Shivakumar · 9 months, 3 weeks ago

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For any integer \(a\), prove that \(a^{37} \equiv a \pmod{1729}\). Svatejas Shivakumar · 9 months, 3 weeks ago

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@Svatejas Shivakumar Its easy just apply keep applying FLT for various factors of 1729 Shrihari B · 9 months, 3 weeks ago

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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. Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 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!! Svatejas Shivakumar · 9 months, 4 weeks ago

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@Svatejas Shivakumar Nicely done. There are a few mistakes in the solution but one may understand the correct part as he reads it. Saarthak Marathe · 9 months, 3 weeks ago

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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? Svatejas Shivakumar · 10 months ago

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@Svatejas Shivakumar 1/2 Saarthak Marathe · 9 months, 4 weeks ago

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@Saarthak Marathe Correct. Can you show your steps? Svatejas Shivakumar · 9 months, 4 weeks ago

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\(a,b,c\) are real numbers such that \(a+b+c=0\) and \(a^2+b^2+c^2=1\). Prove that \(a^{2}b^{2}c^{2}≤\frac{1}{54}\). Akshat Sharda · 10 months ago

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@Akshat Sharda See my note Satyajit Ghosh · 10 months ago

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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. Satyajit Ghosh · 10 months ago

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