At last, nervousness comes down. Exam got over. Thanks to Brilliant, it suited my learning style. I like to learn new stuff through discussion and solving new problems,

Now, here is the paper. I think I have done a decent job

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$</code> ... <code>$</code>...<code>."> Easy Math Editor

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## Comments

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TopNewestLet $\boxed{(x+\frac{1}{x})=m}.......(1)$ . Therefore $(x^{3}+\frac{1}{x^{3}})+3(x+\frac{1}{x})=m^{3}$ $\boxed{(x^{3}+\frac{1}{x^{3}})=m^{3}-3m}.......(2)$ Squaring both sides , we get $\boxed{(x^{6}+\frac{1}{x^{6}})=m^{6}+9m^{2}-6m^{4}-2}........(3)$ Now rewriting the given expression in terms of $m$ we get $=\frac{m^{6}-(m^{6}+9m^{2}-6m^{4}-2)-2}{m^{3}+m^{3}-3m}$ $=\frac{6m^{4}-9m^{2}}{2m^{3}-3m}$ $=3m$ By $A.M-G.M$ $x+\frac{1}{x}\geq2$ Therefore $3m\geq6$ Hence the minimum value of the given expression is $6$

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Did the same

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Question 5 is trivial. Add 1 to the LHS and then it becomes a perfect cube. Then try proving that the RHS is a cube lying between x^3 and (x+2)^3 forcing it to be (x+1)^3. thus u obtain that x=y. Substitute that and get solutions as (1,1),(10,10)

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Perfect solution!

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$\boxed{CP=CD,EP=PF}.....(1)$ (Since $C_{1}P,C_{2}P$ are perpendicular to $CD,EF$ respectively) Angle$CPF$ =Angle$EPF$ Angle$CPE$ =Angle$FPD.....(2)$ Using $(1),(2)$ Triangle CPF is congruent to triangle EPD (BY SAS Test)

Triangle CPE is congruent to triangle FPD (BY SAS Test)

Therefore CF=ED,CE=FD...(I). And Angle$FCD$ =Angle$EDC$ Which implies CF||ED....(3) Similarly we prove CE||FD...(4) Hence using (I),(3),(4) we prove that CEDF is a rectangle.

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What is the cutoff for qualifying to the next round?

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I too want to know the same .... i am nervous about that .. If anyone gets to know it ... PLEASE post it in this discussion ... let the Tamil Nadu people in Brilliant know it ...

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is trichy a centre for RMO

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i hope pic is clear ... else jus reply to me ... i will type out the full question

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3) 80

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Yeah .. I too got the Same

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What method did u use ? mod 100 ?

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That is probably the easiest method.

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$32^{5}=2^{25} \equiv 2^{10} \times 2^{10} \times 2^{5} \equiv 24^{2} \times 32 \equiv 32 \pmod{100}$. Hence last two digits of every fifth power of $32$ is $32$. Hence last two digits of $N$ is last two digits of $2015 \times 32=64480$ which is $80$.

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Is 6(a) 721 and 6(b) 5004?

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Yeah ... I too got the same

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Yeah .I too got the same.

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hey mohan i got 5005 dah

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People .. Any idea abt cutoff?? ..

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Q3) Clearly $N \equiv 0 \equiv 80 \quad (\mod 4)$.

Now we construct a table for $2^{5^{n}} \quad (\mod 25)$.

$\Large{\underline { \begin{matrix} n & & n\quad \left( \mod 25 \right) \end{matrix} } \\ \begin{matrix} 1 & & 7 \\ 2 & & 7 \\ 3 & & 7 \\ & . & \\ & . & \\ & . & \\ & & \end{matrix}}$

This gives $N \equiv 5 \equiv 80 \quad (\mod 25)$. Hence the answer is $\boxed{80}$

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Both your and Svatejas's solutions are good.

What about question 5? How you did?

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I did it like Shrihari B.

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Also, Do you appeared in RMO? From Delhi region?

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I got #1 and #3 right. For #4 I showed that CDEF is a parallelogram since P is the midpoint of CD and EF. Finally for #6a) I got 722 by some arithmetic an error. How much marks would I be approximately getting? Any guesses for the cutoff? Thanks!

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Friends ... The RMO results are out ... Fortunately I hav cleared it and now can write INMO ... but i feel nervous coz i am hardly able to do 1 sum frm previous year papers ... so i wish to get some tips regareding INMO

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Congratulations!

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thanks ... wat about you brother? .. i tink ur region results came before ours ...

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I have got the proof for the 4 th one

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@ Ganesh Ayyappan Hi I am Anurag from Mumbai and currently, I am in 8th standard and selected for RMO.I am able to solve only three or four problems from each RMO paper or even 2 if it is difficult. For sums I am not able to solve, I am able to get through halfway and then I can't solve because I don't the formula or else, I am missing something. Is 3 or 4 problems from 6 alright to get through RMO and go to INMO? Please let me know. Do follow me. I will be readily answering any sum on Brilliant.

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LOL you were damn lucky to get selected.

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The answers i got

1) 6

3) 80

5) x=y=1 & x=y=10

6) A - 721 & B - 5004

i verified once that all are right .. and i tink some of my friends havent noticed that this discussion is put up ....

@Harsh Shrivastava @Adarsh Kumar @Mehul Chaturvedi @Kartik Sharma and some more of my BRILLIANT friends ... awaiting ur valuable replies

Also ... can sum1 post soln of Q2 & 4??

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What do you expect the cutoff to be?

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Since i do not hav any idea .. Im asking here ...

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Diagram : Solution to problem 2. @Ganesh Ayyappan Here is a solution for the problem (2).

$Sol^{n}$:

$1st$ case: When $\angle BDP = \angle CDQ$ then $MP = MQ.$

Constructions: Produce $QM$ to $Q'$ such that $MQ = MQ'.$ Join $BQ' , PQ$ and $PQ'.$

Here clearly, quadrilateral $BQCQ'$ is a $|| gm$. $=>$ $\angle BQ' = CQ$ --- $[1]$ and also, $\angle Q'BM = \angle C$ $=>$ $\angle Q'BP = \angle B +\angle C = 180^{\circ}- \angle A = \angle PDQ$ --- $[2]$

In $\triangle BDP$ and $\Delta CDQ,$ $\angle BDP = \angle CDQ$ and $\angle BPD = \angle CQD$ $=90^{\circ}.$ $=>$ $\Delta BDP \sim \Delta CDQ$ $=>$ $\dfrac{BP}{CQ}= \dfrac{DP}{DQ}$ $=>$ $\dfrac{BP}{BQ'} = \dfrac{DP}{DQ}$ (Using - $[1]$ ) $=>$ $\dfrac{BP}{DP} = \dfrac{BQ'}{DQ}$

Now, in $\Delta BQ'P$ & $\Delta PDQ$ , $\dfrac{BP}{DP} = \dfrac{BQ'}{DQ}$ and $\angle Q'BP = \angle PDQ$ [From $eq^{n}$] - $2$ $=>$ $\Delta BQ'P \sim \Delta PDQ$ $=>$ $\angle BPQ' = \angle DPQ$ $=>$ $\angle BPQ' + \angle Q'PQ = \angle Q'PQ + \angle DPQ$ $=>$ $90^{\circ} = \angle Q'PQ + \angle DPQ = \angle QPQ'$ $=>$ $\Delta QPQ'$ is right angled at $P$.

And we know that the midpoint of the hypotenuse of a right angled triangle is circumcenter of that $\Delta$. $=>$ $MQ = MP$ $= MQ'$

Now, just reverse this process to prove that: If $MP = MQ$ then $\angle BDP = \angle CDQ$.

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Problem 4 is easy. One can easily notice that $EP = FP$ and $DP = CP$ [Because, the $\perp$ from the center to the chord bisects the chord]. => $FDEC$ is a $|| gm.$ Now, since $DACB$ is cyclic, $DP\times CP = AP\times BP = DP^{2}$ -- $1$. Similarly, since $FAEB$ is cyclic, $EP\times FP = AP\times BP = FP^{2}$-- $2$ The two results $1 and 2$ tells that, $DP = FP => 2DP = 2FP => DC = FE$. So, quad $DFCE$ is a $||gm$ with the diagonals equal. => $CEDF$ is a rectangle

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I think this region's paper was the easiest one.

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In fact i found this the toughest paper !

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@Shrihari B I feel that Mumbai region was the hardest. Definitely my region was the easiest but still I couldn't quality that :( Why aren't you participating in the INMO board anyone?

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besides being a 14 yr old ... u are a level 5 solver in all topics u hav took ... tats really appreciable ... so dont worry abt this yr RMO .... GOOD LUCK

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Someone please post solution for q6, it was the easiest but sill want to know how others did it

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Use Principle of Inclusion and Exclusion.

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Got through with part a, finding part b difficult

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