Let $\sigma$ be the sum of divisors function. First I note that If $n \geq 28$ then $\sigma(n) > 28$. So we must have $n < 28$. Next note that if $n$ is a prime then $\sigma(n)=n+1 = 28 \iff n = 27$, which is a contradiction. So $n$ is not a prime. This leaves us to check the numbers $1,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26$. Which the only number that works is $n = 12$, as $\sigma(n)=1+2+3+4+6+12=28$.

But that was a lot of work, can the search be narrowed down even further? Perhaps by looking at what factors $n$ can have? Minimum size of $n$? I know $\sigma(n)\leq n\tau(n)$.

  • 1
    $\begingroup$ Once you have reduced the problem to $n < 28$, just look at a table for $\sigma$, such as oeis.org/A000203/list. $\endgroup$ – lhf May 14 at 16:20
  • $\begingroup$ I guess, but this problem appeared on a past exam, wouldn't have access to table, and wouldnt like to spend precious exam time adding numbers.. $\endgroup$ – pureundergrad May 14 at 16:21
  • $\begingroup$ The sum of divisors function is multiplicative. If you know that and the value for prime powers (easy to derive and remember) you have a good head start. en.wikipedia.org/wiki/Divisor_function#Formulas_at_prime_powers $\endgroup$ – Ethan Bolker May 14 at 16:25
  • $\begingroup$ See for similar questions here and here - and here, etc. $\endgroup$ – Dietrich Burde May 14 at 16:49
  • $\begingroup$ @DietrichBurde , thank you. This was google resistant for me $\endgroup$ – pureundergrad May 14 at 17:18

Let $p$ a prime divisor of $n$, say $n=p^tk$ and $p$ does not divide $k$. Then $\sigma(n)=\sum_{j=0}^tp^j\sigma(k)$.

$\sum_{j=0}^t 2^j$ divides $28$ for $t=2$

$\sum_{j=0}^t 3^t$ divides $28$ for $t=1$

No other divisor of $28$ is a sum of powers of primes.


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.