# 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.

A polite number, it seems, is a positive integer $n$, such that there is a list of consecutive positive integers $a, a+1,\dots, a+r$ with $n = a + (a + 1) + \dots + (a + r)$.

The politeness is the number of representations of a polite number. For example $9$ is polite and its only representations are $2+3+4$ and $4+5$ (as you can verify), so it has politeness $2$.

The politeness of a number turns out to be the number of its odd divisors, greater than one. For example $9 = 3^2$ has the divisors $1,3,9$, the latter two are odd divisors and greater than one, so again: $9$ has politeness $2$. A prime number $p$ has only $1,p$ as divisors, therefore it has politeness $1$ if and only if it is not $2$ (since $p=2$ is not odd).

• thank you! this was helpful – lovelylola Feb 13 '15 at 20:19
• you're welcome. – Stefan Perko Feb 13 '15 at 20:20
• So ... none of the choices given in the question are correct, then? I count five odd divisors greater than one: $(3,5,9,15,45)$. – John Feb 13 '15 at 20:25

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 $(2+1)(1+1)-1=5$.

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 $5$.