# When is $\sigma(n)=28$?

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 $$\sigma(n)\leq n\tau(n)$$.

• Once you have reduced the problem to $n < 28$, just look at a table for $\sigma$, such as oeis.org/A000203/list. – lhf May 14 at 16:20
• 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.. – pureundergrad May 14 at 16:21
• 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 – Ethan Bolker May 14 at 16:25
• See for similar questions here and here - and here, etc. – Dietrich Burde May 14 at 16:49
• @DietrichBurde , thank you. This was google resistant for me – 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.