RMO 2010

GUYS i got the answer of 3 as 738

the no of number having non zero digits and divisible by 4 are 1458 and out of the 720 are divisible by 8 hence 738 is the reqd answer

@rahulmishra how can u compare coefficients alpha is a constant not a variable
@thumbsdown how can u assume angle ABM=angle ABC in ur proof?

2>
alpha is the common root not a.
@rahulmishra
you have proved it incorrectly

http://www.artofproblemsolving.com/Forum/search.php?author_id=66880&sr=posts
this is the link of the posts of the guy who leaked them on a maths forum

ta i too got the answers as

3) 729

4) {2,3,6} : i did this by hit and and trial... ne prob!!!..

6) 43

For 4: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=380707

A SHOCKING EXPERIANCE : -

While following post no. 34 by H . S . Bhatt sir , I actually discovered that the question paper of RMO 2010 in West Bengal Region was leaked in " artofproblemsolving " on 3 December 2010 , whereas the test itself took place on 5 December . You can verify this by clicking the following links -

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=380959&start=0
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=380700&start=0
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=380828&start=0
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=58&t=380708&start=0
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=380706&start=0
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=380702&start=0

Now , I have sent an e - mail to the Joint Co - ordinator in this context . What do you think they should do ?

When I saw Qn #2, I was scratching my head bcoz I had just recently seen it and I could not recollect where :D

Turns out i saw the leaked question!

that may not be correct (3,4,12) is a counterexample. My proof shows that all solutions are of the form (2k,3k,6k)

yes sir ...the ratio only matters.....

wow questions wer leaked...i did 4 correct...now whats gonna happen??

Q4 wasnt all about hit and trial........... but it was easy if ur brain cells helpd u!!!
The answer was {2,3,6} rt?

Is solving 3 questions gud for a class 10th student????

guys,

i present here the solution 0f q no 5
please tell if it is right or wrong

we had to prove that m lies on BC
that is easy to prove
if we prove <amb +<amc = 180 we are done

in tri AMB, AMB + BAM + ABM = 180
in tri AMC, AMC + CAM + ACM = 180

adding we get, AMB + AMC + 60 + (180 - 60) = 360
=> AMB + AMC = 180
=> B,C,M ARE COLLINEAR

@ishita.. -> Check out pioneermathematics.com, they've uploaded the soln.

@Seniors -> Please tell me whr am i wrong in my soln.......

if, ap² + bp + c = 0 then why can't a = 0, b = 0 and c = 0 where p is a constant...!

@rahulmishra
a, b and c need not be zero eg alpha=1, a=2, b=-1, c=-1

Q2. Easiest way-

P₁=P₂=P₃

a\alpha ^{2}-b\alpha -c = b\alpha ^{2}-c\alpha -a=c\alpha ^{2}-a\alpha -b
\Rightarrow a\alpha ^{2}+a\alpha +a= b\alpha ^{2}+b\alpha +b= c\alpha ^{2}+c\alpha +c

cancelling \alpha ^{2}+\alpha +1 from both sides we get

a=b=c

I saw sum site which displayed, "RMO 2010 questions leaked.........but RMO not being conducted again!"

you cannot know

they dont give out the marks as far as i know

I know that was wrong and but when i wanted to delete it ,it was too late
Anyway I think that this might be the right proof
Since P₁=P₂=P₃
(a-b)x²+(c-b)x+(a-c)=0 - 1
(b-c)x²+(a-c)x+(b-a)=0 - 2
(a-c)x²+(a-b)x+(b-c)=0 - 3
Hence
From 1 & 2 we get

-(c-b)+/-√(b-c)²-4(a-b)(a-c)2(a-b)=-(a-c)+/-√(a-c)²-4(b-a)(b-c)2(b-c)

Therefore, (b-c)²-4(a-b)(a-c)=(a-c)²- 4(b-a)(b-c)
From that we get
(b-a)(a+b-2c)=0
Therefore
a=b
or,
a=2c-b

If we do the same for 1 and 3 we get

a=c
or,
a=2b-c
from here we can easily derive
a=b=c

3) sum 1line answer
1st asume no a=b=c
whatever be the root the discrimintant of all the equations will be same as 1 root is same for all
thereofre
b^2+4ac=c^2+4ab
solving we get
(b-c)(b+c)=4a(b-c) 1
therefore b+c=4a
similarly a+b=4c
a+c=4a
thereofre therein lies some flaw bcoz on adding it we get a+b+c=0

thereofre eq 1 2 or 3 can only be wrong when
either
a-b,b-c,a-c=0 thereofre we assume any of them anfd prove it

Q. 3 > 881

Q . 4 > You can easily see that it is not a solution . The correct solution is - { 2 , 3 , 6 }

Q . 6 > 43

Proof for Qs . 3 >

Divide the numbers into groups of hundreds .

In each group , there are exactly 11 numbers that satisfy the given conditions .

Hence , the answer = 9 x 9 x 11 = 881