# How many even positive integers are there that are divisors of 720?

How many even positive integers are there that are divisors of 720 ?

I know how to compute the number of divisors but how to compute the number of even or odd positive divisors of a number ?

If we list the divisors of 720 (using mathematica) : {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, 40, 45,48, 60, 72, 80, 90, 120, 144, 180, 240, 360, 720} among these only 24 are even,I am looking for some tricks that can be used in solving similar kinds of problems during exam (under a minute solution).

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If you already know how to compute the number of divisors, you can use this. –  Guess who it is. Oct 29 '10 at 10:26
@J.M: I think it is giving The sum of powers of even divisors of a number this is not what I want,even I have no idea if that can be used to compute the number or even or odd divisors ?! –  Quixotic Oct 29 '10 at 10:28
...and a number raised to the zeroth power is? –  Guess who it is. Oct 29 '10 at 10:29
..and a number raised to the zeroth power is? I believe it's 1. –  Quixotic Oct 29 '10 at 10:31

An even number $2m$ is a factor of $720$ iff $m$ is a factor of $360$. So it's the same problem as counting divisors of $360$. For that it helps to consider the prime factorization of $360$.

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Let the prime factorization of the integer $n$ be $n=\prod_{i=1}^{k}p_{i}^{e_{i}}$. The number of divisors of $n$ is given by $$\prod_{i=1}^{k}\left( e_{i}+1\right) .$$ For $n=720$, we have $$720=\prod_{i=1}^{3}p_{i}^{e_{i}}=2^{4}\times 3^{2}\times 5$$ and $$\frac{n}{2}=\frac{720}{2}=360=2^{3}\times 3^{2}\times 5.$$ Hence the number of divisors of $360$ is

$$\prod_{i=1}^{3}\left( e_{i}+1\right) =(3+1)(2+1)(1+1)=24.$$

All these divisors of $n/2=360$ are even divisors of $n=720$ and there are no more even divisors of $720$. So there are $24$ even divisors of $720$.

ADDED. In response to Wishingwell's comment here is a possible proof. Every divisor of $n=\prod_{i=1}^{k}p_{i}^{e_{i}}$ is of the form $$\begin{equation*} \prod_{i=1}^{k}p_{i}^{a_{i}},\qquad 0\leq a_{i}\leq e_{i} \end{equation*}$$

and a term of the product $$\begin{equation*} \underset{e_{1}+1\text{ terms}}{P=\underbrace{\sum_{j=0}^{e_{1}}p_{1}^{j}}} \times \underset{e_{2}+1\text{ terms}}{\underbrace{ \sum_{j=0}^{e_{2}}p_{2}^{j}}}\times \cdots \times \underset{e_{k}+1\text{ terms}}{\underbrace{\sum_{j=0}^{e_{k}}p_{k}^{j}}}. \end{equation*}$$

Since each sum has $e_{i}+1$ terms $(1\leq i\leq k)$, the number of terms of $P$ is $\sum_{i=1}^{k}(e_{i}+1)$.

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I'd be interested in seeing a combinatorial argument that gives the sum you have written. Just a curiosity! –  Starlight Jun 19 '13 at 4:32
@Wishingwell Every divisor of $n=\prod_{i=1}^{k}p_{i}^{e_{i}}$ is of the form \begin{equation*} \prod_{i=1}^{k}p_{i}^{a_{i}},\qquad 0\leq a_{i}\leq e_{i} \end{equation*} and a term of the product \begin{equation*} \underset{e_{1}+1\text{ terms}}{P=\underbrace{\sum_{j=0}^{e_{1}}p_{1}^{j}}} \times \underset{e_{2}+1\text{ terms}}{\underbrace{ \sum_{j=0}^{e_{2}}p_{2}^{j}}}\times \cdots \times \underset{e_{k}+1\text{ terms}}{\underbrace{\sum_{j=0}^{e_{k}}p_{k}^{j}}}. \end{equation*} Since each sum has $e_{i}+1$ terms $(1\leq i\leq k)$, the number of terms of $P$ is $\sum_{i=1}^{k}(e_{i}+1)$. –  Américo Tavares Jun 19 '13 at 9:23

There is a very simple trick for this,first compute the prime factorization of $720$,which is $2^4 \times 3^2 \times 5$,the total number of factors here is $3 \times 2 \times 5 = 30$, and number of odd factors (number of factors of the odd primes)$=3 \times 2 = 6$,subtracting gives number of even factors = $24$.This method works for any number.

NOTE: If the number has no odd factors i.e,the prime factorization is of the form $2^a$,then the number of number of even factors is $a$ and number of odd factors is $1$.

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You have an even number given, so just cut it in half and count the divisors of this.

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If n is a divisor of 360, then 2n is an even divisor of 720 and vice versa. So there are the same number in each set. –  Ross Millikan Oct 29 '10 at 13:59

1*2*2*2*2*3*3*5. Odd divisors: (1,3,3,5) => (1,3,5,9,15,45). Let A = (1,3,5,9,15,45). Then all factors are, A, 2*A, 4*A, 8*A, 16*A.

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Could you please explain in a bit elaborately ? –  Quixotic Oct 29 '10 at 10:48