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.

As per Wikipedia, a polite number is one that can be written as sum of consecutive integers, and its politeness is the number of such representations. Looking at such a representation
$$ 90=n+(n+1)+\ldots +(n+m)$$
we see from the familiar formula for arithmetic series that this can be rewritten as 
$$ 90 = (m+1)\cdot n + \frac{m(m+1)}{2}=\frac{(m+1)(m+2n)}2$$
Note that exactly one of $m+1, m+2n$ is odd, so each polite representation of $90$ gives rise to an odd divisor of $90$.
On the other hand, if $d$ is an odd divisor of $90$, we can attempt either


*

*$m=d-1$, $n=\left(\frac{2\cdot90}{m+1}-m\right)/2=\frac{90}d-\frac{d-1}2$

*or $m=\frac {2\cdot 90}d-1$, $n=\frac{d-m}2$ 


For the first method to work, we need $d>1$ and $\frac{90}d>\frac{d-1}2$, i.e., 
$\frac{2\cdot 90}d >d-1$.
For the second method to work we need $d>m$, i.e., $ \frac{2\cdot 90}d<d+1$. Since $\frac{2\cdot 90}d$ is even, exactly one of the options works for each odd divisor $d>1$.
Therefore the politeness is the number of odd divisors. This can be read from the prime factorization of $90=2\cdot 3^2\cdot 5^1$: the odd divisors are of the form $3^a5^b$ with $0\le a\le 2, 0\le b\le 1$. Hence there are $3\cdot 2$ odd divisors. Remove $d=1$ to arrive at politeness $5$.
