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

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    $\begingroup$ 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
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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.

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