Famously, a perfect number is equal to the sum of its proper divisors. For example, 28 is equal to 1 + 2 + 4 + 7 + 14. If the sum is more than the original, the number is called abundant, and if the sum is less, the number is called deficient. For example, 12 is abundant, since 1 + 2 + 3 + 4 + 6 = 16, which is 33% more than 12.

It's clear numbers can become arbitrarily more and more deficient, but abundance is a bit trickier.

Specifically, is there a number whose proper divisors sum to more than ten times the original number? Why or why not? The puzzle is to come up with a short mathematical argument resolving this question.

My spouse came up with this puzzle during a trip to Dairy Queen, so it's "original" in a sense but these concepts are well studied and I'm sure the question and its answer appear somewhere.


4 Answers 4


In short:

Numbers can get as abundant as you like.

Consider factorials.
N! is divisible by 2,3,..., N by definition.
So all of N!/2, N!/3, ..., N!/N are also factors.
So ignoring other factors, we already have a sum of N!/2 + N!/3 + ... + N!/N
= N!*(1/2 + 1/3 + ... + 1/N)
= N!*(The first N terms of the harmonic series - 1)

The harmonic series is proven to be divergent, so for suitably large N, you can find a number of arbitrary abundance.

To answer the specific question asked, the abundance factor passes ten somewhere around 12400!. (There may of course be smaller numbers that meet this requirement.)

  • 1
    $\begingroup$ Notably, you can instead consider the LCM of all these numbers, which will still be large, but not quite as large. $\endgroup$ Commented Jul 18, 2023 at 22:36
  • 2
    $\begingroup$ This is the intended solution, nice work! $\endgroup$ Commented Jul 19, 2023 at 1:08

Does such a number exist?

Short answer: Yes; the Multiply Perfect Numbers page claims there are two whose ratio is exactly ten, and so any multiple of either will have a larger ratio. The rest of this answer is the "long answer".

We require a few priors:

The series $\sum\limits_{p\text{ prime}}\frac{1}{p}$ diverges. This is a well-known result dating back to Euler.
The sum of divisors $\sigma(n)$ of $n$, and its ratio $\frac{\sigma(n)}{n}$, are multiplicative functions. This is because divisors of n correspond to choices of exponent for each of n's prime factors, and every such set of choices is represented; thus, we can collect each possible power of a given prime into a sum, multiply those sums together, and the process of multiplication will achieve the same process of finding every divisor. Additionally, we can divide each of those sums by the highest powers of their respective primes to get a product form for $\frac{\sigma(n)}{n}$.

Where to begin?

If we look at the product form for $\frac{\sigma(n)}{n}$, each prime can contribute a factor less than $1+\frac{1}{p-1}$. Importantly, by making $n$ a multiple of a sufficiently high power of $p$, $p$'s contribution can be arbitrarily close to this limit. We can see that $1+\frac{1}{p-1} > e^{\frac{1}{p}}$ by expanding both sides as power series in $\frac{1}{p}$, and since $\sum\limits_{p\text{ prime}}\frac{1}{p}$ diverges, we can choose some sufficiently large cutoff $N$ such that $\sum\limits_{p\text{ prime}}^N\frac{1}{p} > \ln(11)$, so that $\frac{\sigma(n)}{n} > \Pi\left(1+\frac{1-\frac{1}{p^k}}{p-1}\right) > e^{\left(\sum\limits_{p\text{ prime}}^N\frac{1}{p}\right)} > 11$, and we are done.

P.S. Why 11? Because working with the sum of divisors is easier than working with the abundance, and if the abundance is ten times the number, then the sum of divisors is eleven times the number.

  • 2
    $\begingroup$ Upvoting this answer because it provides a rigorous mathematical proof with elaboration unlike the other answers. $\endgroup$ Commented Jul 18, 2023 at 22:34

It must be yes N! is divisible by 2,3,… N And so therefore N!/2, N!/3 etc are also all factors. The harmonic series diverges and hence the result


Do such numbers exist?

Yes, there is a number n for which the sum of the proper divisors of n is greater than 10n, or equivalently, the sum of all divisors of n is greater than 11n.

Here's why:

For any positive integer n, the sum of all divisors of n divided by n is the same as the sum of the reciprocals of the divisors of n. The harmonic series diverges, so it will eventually reach a sum greater than 11 (or in fact any finite value). The first time this happens is at the 33,617th partial sum, so if we let n be the lcm of all positive integers from 1 to 33,617, then the sum of all proper divisors of n is greater than 10n.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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