How many ways can 720 be expressed as a sum of n Consecutive Integers

We can use the property of sum of n consecutive integers=n*average of the numbers.
Also, average of consecutive integers is the middle or median of the numbers.

Average when n is even
The average of the integers must be a decimal of the form xy.5, as it is middle of two consecutive numbers.
And when do we have that?: When we have just one 2 extra in the divisor
\(720=2^4*3^2*5\)
1) We have \(2^4\) in 720, so \(2^5\) will get a decimal of the form xy.5
\(\frac{720}{2^5}=22.5\)
32*22.5….n=32
7,8,9,….,21,22,23,24,…..36,37,38

2) We have \(2^4*3^2\) in 720, so \(2^5*3^2\) will get a decimal of the form xy.5
\(\frac{720}{2^5*3^2}=2.5\)
But this will get negative integers too.

3) We have \(5*2^4\) in 720, so \(5*2^5\) will get a decimal of the form xy.5
\(\frac{720}{5*2^5}=4.5\)
But this too will get negative integers.

Average when n is odd
The average of the integers will be an integer.
And when do we have that?: When we divide by an odd factor.
\(720=2^4*3^2*5\)
1) We have \(3\) in 720, so
\(\frac{720}{3}=240\)
3*240….n=3
239,240,241

2) We have \(3^2\) in 720, so
\(\frac{720}{3^2}=80\)
9*80….n=9
76,77,78,79,80,81,82,83,84

3) We have \(5\) in 720, so
\(\frac{720}{5}=144\)
5*144….n=5
142,143,144,145,146

4) We have \(3*5\) in 720, so
\(\frac{720}{3*5}=48\)
15*48….n=15
41, 42,….48,…..54, 55

5) We have \(3^2*5\) in 720, so
\(\frac{720}{3^2*5}=16\)
16*45….n=45
If average is 16, we will require 22 integers lesser than 16, and it will result in negative integers too.

Thus, total 5