The number $90$ is a polite number, what is its politeness? 
The number $90$ is a polite number, what is its politeness?
A. $12$
B. $9$
C. $6$
D. $14$
E. $3$

How did you get that answer? I tried Wikipedia to figure out what a polite number was and how to figure out its politeness but I'd like to see it done step by step or have it explained because I just don't understand.

From Wikipedia

In number theory, a polite number is a positive integer that can be written as the sum of two or more consecutive positive integers.
The politeness of a positive number is defined as the number of ways it can be expressed as the sum of consecutive integers.

You have the following result:

For every $x$, the politeness of $x$ equals the number of odd divisors of x that are greater than one. 

As idea of proof, if you have a writing of $x$ as $m+(m+1)+\dots +(m+k)$, where $m, k\geq 1$ its sum is equal to $m\cdot (k+1)+k(k+1)/2=(k+1)(2m+k)/2$. So you have to have $(k+1)(2m+k)/2=x$. One of the two $k+1$ or $2m+k$ is odd and they are both greater than $1$, so each writing corresponds to a odd divisor. 
The other way round, if you have an odd divisor, $y$, then you have the writing: $x=\sum_{i=\frac{x}{y} - \frac{y-1}{2}}^{\frac{x}{y} + \frac{y-1}{2}}i.$
Some of the terms in this sum may be zero or negative. However, if a term is zero it can be omitted and any negative terms may be used to cancel positive ones, leading to a polite representation for $x$. (The requirement that $y>1$ corresponds to the requirement that a polite representation have more than one term). (Wikipedia)
The number of divisors of a number $x:=a_1^{b_1}\dots a_n^{b_n}$ is given by the formula: $(b_1+1)\dots (b_n+1)$. Since you are interested only on the odd divisors greater than 1, you will ignore the power of $2$ and you will subtract $1$ (because you do not want to take $1$ into consideration). For $90=2^1\times 3^2\times 5$, you will have $(2+1)(1+1)-1=5$.
