How Many Consecutive Integers Produces This Sum?

Assume that S can be expressed as n consecutive integers.

Then S = nk + $\frac{n(n+1)}{2}$ where k is an integer.

If n is odd then S will be divisible by n .

5 ! is divisible by only 3 odd numbers 3 , 5 , 15 . these are the only odd possible n .

If n is even then S = nk + $\frac{n(n+1)}{2}$ , so S must be devisible by $\frac{n}{2}$ but S can not be devided by n .Here we will take $\frac{n}{2}$ even otherwise it itself will be a possible n .

the even divisors of 5 ! are 2 , 4 , 6 , 8 , 10 , 12 , 20 , 24 , 30 , 40 , 60 , 120.

If we take 2 as $\frac{n}{2}$ then n will be 4 but 4 is a divisor of 5 ! .So we will eliminate 2 from our possiblity.

Thus , we will eliminate 4 , 6 , 10 , 12 , 20 , 30 , 60.

we can only take 8, 24 , 40 ,120 as $\frac{n}{2}$ for even $\frac{n}{2}$

So the possible even n(s) are 16 , 48 , 80 , 240.

So the sum of all possible n(s) are 3+5+15+16+48+80+240 = 407

$\text {Let } a \in \mathbb {Z} \text { be the first term of this arithmetic sequence (common difference = 1). }$

Then:

$S = \frac {2a + (n - 1)}{2} × n = 120$

$(2a + (n - 1)) × n = 240$

Since 2a is always even ( and even + even = even; even + odd = odd) and (n-1) and n are consecutive integers, therefore one of these factors of 240 has to be odd.

The odd factors of 240 : 1, 3, 5 and 15

Their even "pairs" (with which they make a product of 240):

240, 80, 48 and 16, respectively.

Since "the sum of consecutive integers" indicates n > 1 , therefore the possible values of n (it is easy to see that we can get integers for a, when we substitute these values for n):

3, 5, 15, 240, 80, 48 and 16.

Hence, the sum of all possible values of n is:

$3 + 5 + 15 + 240 + 80 + 48 + 16 = \boxed {407}$