<s>[INST] Determine in How many ways $N!$ can be expressed as sum of consecutive numbers Determine in How many ways $N!$ can be expressed as sum of consecutive numbers.For example for $3!$ can be as $1+2+3$ only $1$ ways.
Please suggest an elegant approach.
 [/INST] I am assuming that you want it to be written as a sum of consecutive natural numbers.
Let $p \leq N$ be a prime. The highest power of $p$ dividing $N$ is given by $\alpha_p = \displaystyle \sum_{k=1}^{\infty} \lfloor \frac{N}{p^k} \rfloor$.
So $N! = 2^{\alpha_2} 3^{\alpha_3} 5^{\alpha_5} \ldots q^{\alpha_q}$, where $q$ is the last prime $\leq N$.
We want $N! = a + (a+1) + (a+2) + (a+3) + \cdots b = \frac{(b-a+1)(a+b)}{2}$, where $a,b \in \mathbb{N}$
So we need to find $a$ and $b$ such that $(b-a+1)(a+b) = 2^{(\alpha_2 + 1)} 3^{\alpha_3} 5^{\alpha_5} \ldots q^{\alpha_q}$.
Now note that $(a+b)$ and $(b-a+1)$ are of opposite parity.
So one of them has to be odd. Further $(a+b)>(b-a+1)$ since $a \in \mathbb{N}$.
Assume that $(a+b)$ is even. So $(b-a+1)$ is odd.
Hence 
$$(a+b) = 2^{(\alpha_2 + 1)} 3^{\beta_3} 5^{\beta_5} \ldots q^{\beta_q}$$ $$(b-a+1) = 3^{\alpha_3-\beta_3} 5^{\alpha_5-\beta_5} \ldots q^{\alpha_q-\beta_q}$$
where $0 \leq \beta_p \leq \alpha_p$ and $(a+b) > (b-a+1)$
Now assume that $(a+b)$ is odd. So $(b-a+1)$ is even.
Hence 
$$(a+b) = 3^{\beta_3} 5^{\beta_5} \ldots q^{\beta_q}$$ $$(b-a+1) = 2^{(\alpha_2 + 1)} 3^{\alpha_3-\beta_3} 5^{\alpha_5-\beta_5} \ldots q^{\alpha_q-\beta_q}$$
where $0 \leq \beta_p \leq \alpha_p$ and $(a+b) > (b-a+1)$
If we relax the fact that it has to be written as a sum of consecutive natural numbers, and assume that the consecutive numbers belong to integers then we get
$$2 \times (1+\alpha_3) \times (1+\alpha_5) \times (1+\alpha_7) \cdots \times (1+\alpha_q)$$
Note that the above also acts as a trivial upper bound if it has to be written as a sum of consecutive natural numbers. This upper bound is obtained by violating the constraint $(a+b) > (b-a+1)$
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