This was "disappointingly " easy for RMO .... very easy .

Let's see the amount of times where that number is divisible by 9. First off, we know there is MAXIMUM two three's, so let's factor that in. The rest of the digits sum to 3 mod 9. We also know that 1 and 2 are the only other digits that can be in the number, and there must be maximum four total, so by constructing bounds(1111=4 and 2222=8) we know that there is no possibility that satisfies the restraint that the number must be divisible by 9. kk108 wrote: This was "disappointingly " easy for RMO .... very easy . Yes. I can't make AIME and I can solve it(now did I silly?). It looks like a mid-AMC 10. I have no idea why I'm on here. Good buy!

PLEASE SOMEONE GIVE A DETAILED SOLUTION...I FACE DIFFICULTY IN COMBINATORICS

For a number to be divisible 9, we must have the sum of the digits divisible by 9. let the number contain digits a,b,c,d,3,3. The sum is a+b+c+d+6. thus we can say a+b+c+d=3,12 (for the total sum to be divisible by 9) but since a,b,c,d=1 or 2 , 4<a+b+c+d<8 hence we can conclude there exist no such numbers By the way, this questions require no combinatrics

Can anyone post the paper

So sad to see this question in rmo. I think there be some error.

Check out the official solution if you have any query I am providing the link [url][/url]http://olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/11/3sol.pdf Solution

Lemma If $a \equiv b \pmod{r}$ for $a,b,r\in \mathbb{Z}$. Then for any polynomial $P$with integer cofficient $P(a)\equiv P(b) \pmod{r}$.. This lemma can be proved very easily By this lemma we can prove that a natural number is divisible by $9$ iff sum of it's digits are also divisible by $9$ Suppose another four digits of the number are$a,b,c,d$ according to the question $a,b,c,d\in \{1,2\}$. So,$9| 6+a+b+c+d$ . No such configuration is possible this can be seen by just looking into one case.

Claim No such number exists. FTSOC, Let $N$ denote such a number. Let $x$ be the number of $1$'s and $y$ the number of $2$'s in $N$ where $x,y \in \mathbb{Z^+}$ Now notice that $2x+y+6\equiv 0\pmod 9$ and, $x+y=4\equiv 4\pmod 9$ Subtracting these modular equations give, $x\equiv -10\equiv 8\pmod 9$ But $x \in \mathbb{Z^+}$ and $x \leq 4$. This is a contradiction. Our proof is thus complete.