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I've been given of this problem: Let $r$ be an integer which has $k$ even divisors and $k-3$ odd divisors. Furthemore let $x$ denote sum of all even divisors and $y$ sum of all odd divisors. What are the all possible values of ratio $\frac{x}{y}$?

I've come to some ideas, which lie in writing $r$ as the following fraction:$$r=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_k^{\alpha_k},$$where $p_1\neq p_2\neq\ldots\neq p_k$ (symbol $p$ is denoting primes). Then for the some divisor $d$ we have $$d=p_1^{\beta_1}p_2^{\beta_2}\ldots p_k^{\beta_k},$$where $$0\le\beta_k\le\alpha_k$$ Number of all divisors of $r$ is then $$(\alpha_1+1)(\alpha_2+2)\ldots(\alpha_k+k)$$

Has any of you got any possible ideas how to solve this problem? Thanks a lot.

Mr. Numbers
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  • Hint: The total number of divisors of a number is odd if and only if the number is square. Show that thus $n$ is a square, and therefore that $k-3$ must be odd. – Thomas Andrews Jan 10 '16 at 23:55

2 Answers2

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First write $r$ as a product of prime factors, say $r=2^a3^b5^c...$. Now you can find the divisors by using the prime factorization. Obviously anything even will have at least one 2, and everything odd with have no 2's.

The restriction that there be 3 more even divisors than odd turns out to be a very rigid one. Let $d$ denote the number of odd divisors. You can then see that the number of even divisors is $ad$. So when is $ad-d=(a-1)d=3$ for $a,d \in \mathbb{Z}$? We have two cases $a=4,d=1$ or $a=2,d=3$

Since 1 is odd our first case gives us $r=2^4$.

What about the second case? You can show that $d>3$ if you have two (or more) odd primes in your prime factorization. Say they are $p,q$, then $1,p,q,pq$ are odd divisors. So we must have $r=2^4p^n$ for $p$ an odd prime. What is $n$? You should be able to show $n=2$ without too much hassle.

I trust you can find the ratios from here.

siegehalver
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Ignoring the $k$, for every positive power of $2$ in the prime factorization, there is an even factor of $n$ that corresponds to an odd factor of $n$. More specifically, you can derive that $x=2y + 4y + \cdots 2^r y$, where $r$ is the power of $2$ in the prime factorization of $n$. The ratio falls out.

Now considering $k$, note that there are exactly $r$ times as many even divisors as odd divisors. Thus $k = r (k-3)$ or equivalently $k(\frac{r-1}{r}) = 3$, so $r|k$. Let $a = k/r$, so $a(r-1) = 3$. Thus either $r=2$ (and $a=3$) or $r=4$ (and $a=1$), making the ratio $x/y$ either $2+4=6$ or $2+4+8+16=30$.

Bob Krueger
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