RMO2 paper

the 4th question can be done uses the cases of even nos in that 4 digit no...

Case 1:If there is 1,3,5 and 0
total cases:3 x 3 x 2 x 1 = 18 cases

Case 2:If there is 1,3,5,(2/4)
total cases:2 x 4!=48

Case 3:two odd nos and a 0 and a even no....
3C2 x 2C1 x 2 x 4 = 48(as de even no can be placed in 4 places)

Case 4: two odd nos and 2 even nos except 0
4! - 3C2 x3 x 2 =6

total cases:18+48+48+6=120

correct solution for 4 ??

2. problem 2 was pretty simple,

we have ,
2n+1=a²
3n+1=b²
as a² is odd it is of the form 8t+1 for some int t.
then we get n=4t,
this implies 3(4t) +1 is odd so b² is also odd.
as both a & b are odd perfect squares so they must be of the form
8k+1.
so,
2n+1=8l+1,
3n+1=8m+1,
for some l,m
form there we get,
n=4l &
n=8m/3

this implies 3|n & 8|n
so n is of the form
24j {obviously,j≠0}
now ,
5n+3=5(24j)+3=
3(40j+1)
which is clearly composite!!

Subho rocks.! :)

Wonderful solution indeed.

1)

The fact I'm going to use without prove is that reflection of H on the line BC lies on the circumcircle of \triangle ABC.

Let H_1 be the reflection of H on the line BC. Let HH_1 intersect BC at T

Join AO, let it meet the circum circle at X. Join HX. Let it intersect BC at M. Equivalently we have to prove that M is the mid point of BC.

Since \angle AH_1X = \frac {\pi}{2}, BC \parallel H_1X. Since \triangle HMT \cong \triangle H_1MT, HM= H_1M = MX.

Now consider \triangle BHM and \triangle XMC.

\angle HBM = \angle XCM = \frac{\pi}{2} - \angle ACB

\angle HMB = \angle XMC

HM= MX

So \triangle BHM \cong \triangle XMC

So BM= MC , QED

plz elaborate!!!!

@ ketan there are 150 permutations

2. Let 2n+1=k² and 3n+1=l².
So, 5n+1=(2k+l)(2k-l)
Claim: 2k-l>1...(1)
Proof:Putting k=√(2n+1 and l=√(3n+1) in (1) we get,
2√(2n+1)>1+√(3n+1)
Squaring we get,
8n+4>1+(3n+1)+2√(3n+1)
→5n+2>2√(3n+1)
Now, (5n+2)/2>√(10n)>√(3n+1)
So, our claim is true and 5n+3 is composite

this paper was simpler than rmo 1

@ shubhodip why is my proof incomplete??

@ vivek it would have been gud if had seen the solution......i would have got 19

marks more

for the 2:

2n +1 = a^2

3n + 1 = b^2

eliminating n;

1 = 3 * a^2 - 2* b^2

so 5n + 3 = a^2 + b^2+ 3* a^2 - 2 b^2

= 4 * a^2 - b^2

= 2a- b * 2a + b

so 5n + 3 is composite

Ohh Wow.. that last question of mine.. great!

http://www.targetiit.com/iit-jee-forum/posts/find-largest-n-20203.html

same question came in de exam.....