In how many ways 1387 can written in the sum of $n,(n>2)$ Consecutive natural numbers In how many ways $1387$ can written in the sum of $n(n>2)$ Consecutive natural numbers?

1.$2$
2.$3$
3.$4$
4.$5$

First we can see that it can be written in the form of the sum of two Consecutive natural numbers.Try other cases.The answer is going to be $3$ but how can we prove it?If you find all three cases how can we be sure that there isn't any other?
 A: HINT:
If $a$ is the first term of the $n$ consecutive natural numbers,
we have $$\dfrac n2\{2a+(n-1)\}=1387$$
$$\iff n^2+(2a-1)n-2\cdot1387=0$$
As $n$ is a natural number, the discriminant $(2a-1)^2+8\cdot1387$ has to be perfect square
Let $(2a-1)^2+8\cdot1387=(2b+1)^2$ where integer $a\ge1,2b+1>\sqrt{8\cdot1387}$
$\iff(b+a)(b-a+1)=2\cdot1387=2\cdot19\cdot73$
So, $b+a$ must divide $2\cdot19\cdot73$
A: If $1387$ is the sum of the $n$ natural numbers $a,a+1,a+2,\ldots, a+n-1$, then $$1387 = \frac{n\cdot(2a+n-1)}2 $$
and hence it suffices to compare all factorizations of $2774$ with the factorization $n\cdot (2a+n-1)$
A: Hint:
Let the consecutive nature numbers be $m,m+1,m+2...m+n$ where $n \geq 0$
The sum of these nature numbers ,by arithmetic progression summation formula, should be $(n+1)(m+m+n)/2 = (n+1)(2m+n)/2 = 1387$
=>
$(n+1)(2m+n) = 2774 = 2 \times 19 \times 73$
Because $2m+n = 2774/(n+1) \geq n \iff n+1 \leq 53$
=> $n+1 \in \{2,19,38\}$
When $n = 1$ it doesn't satisfy the condition so it can be ignored.
When $n = 18, m = 64$ one solution.
When $n = 37, m = 18$ another solution. 
A: The sum of three consecutive numbers is always a multiple of 3.  For example, $6+7+8=3\times7$.  What about the sum of five consecutive numbers, or seven?
On the other hand, the sum of 4 consecutive numbers is a multiple of 2, because $6+7+8+9=(6+9)+(7+8)=2(6+9)$.  What about six consecutive numbers, or eight?
A: The sum of the numbers from m to k is the number of terms, k-m+1 times the average of m and k, (m+k)/2, giving us (k - m + 1)(m+k)/2 = 1387 or (k - m + 1)(m + k) = 2774.  Factor 2774 = 2 x 19 x 73.
If we set n = (k - m + 1) and p = (m + k), then  m = (p - n + 1)/2, where n and p are factors of 2774 with p > n.  The only factorizations we can use are 73 x 38 and 146 x 19.  This gives m=18 and n = 38 and m=64 and n=19.
