If $n$ has a prime factor $p_{k+1}$ which is different with $p_i (0\le p_i\le k)$, contradiction with modular. So $n$ has only $2, p_1, p_2, \ldots, p_k$ as prime factors. If $n=2^m (0\le m\le a)$ then $\frac{n(n+1)}{2}\equiv2^{m-1} \pmod{2^m}$ but $N\equiv0 \pmod{2^m}$, contradiction. If $n$ is the divisor of $\frac{N}{2^a}$, it satisfies the second condition. (This can be proved easily) But there can be a certain $n$ which doesn't satisfy the first condition. Let $n=\frac{N}{2^ar}$. If $n$ doesn't satisfy the first condition, $n=2^{a+1}r$ does. Q.E.D.

This could be full proof?

cloneofsimo wrote: This could be full proof? I wrote this shortly, because I didn't have time to write here. I wrote the full solution in test.

well I managed to prove that $a$ doesn't effect to the number of solutions but actual proof was cumbersome...

Skravin wrote: well I managed to prove that $a$ doesn't effect to the number of solutions but actual proof was cumbersome... If you think of $n=2^{a+1}r$ then easy. I think this was easier than the P1 NT

Skravin wrote: well I managed to prove that $a$ doesn't effect to the number of solutions but actual proof was cumbersome... But cumbersome... I agree.

My solution from the actual contest. Basically, $2N-n(n+1)$ is a multiple of $2n$. Let's ignore the first condition for now. If $n$ is odd, then the condition is equivalent with $n|N$. If $n$ is even, then $v_2(2N) = v_2(n)$, so $n|2N$ but $n \not| N$. Therefore, our candidates for $n$ are divisors of $2N$, with $v_2(n)=0$ or $v_2(n)=v_2(2N)$. So this motivates swapping - let $s(N) = \frac{N}{2^a}$. Take any divisor $x$ of $s(N)$. There are $\prod_{i=1}^k (b_i+1)$ of them. The possible candidates for $n$ are $x$ and $\frac{2N}{x}$. It is easy to prove that exactly one of these two is in our range for $n$. Therefore, the number of possible $n$ is $\prod_{i=1}^k (b_i+1)$, as required.