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).

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.