27. Find all perfect numbers whose prime factorization includes each prime to an odd power. (Recall that a natural number is called perfect if it is equal to the sum of all its natural divisors, less than the number itself - for example, $28=1+2+4+7+14$.)
$\square$ Answer: $n=6$.

Let $p$ be the smallest prime divisor of our number $n$, and $2k+1$ be the power of its occurrence in the prime factorization of $n$. We can divide all divisors of $n$ into pairs $(x, px), (p^2x, p^3x), \ldots, (p^{2k}x, p^{2k+1}x)$, where $x$ is a divisor not divisible by $p$. The sum in each pair is divisible by $p+1$, and thus the total sum of divisors, which is $2n$, is also divisible by $p+1$. This is only possible if $p=2$ (otherwise, $p+1$ is coprime with $n$ due to the minimality of $p$), so $n$ is divisible by $p=2$ and $p+1=3$, i.e., $n=6k$. If $k>1$, then the sum of the divisors of $n$ is at least $1+k+2k+3k+6k=12k+1>2n$—a contradiction. Therefore, $k=1$ and $n=6$. It is clear that 6 works. 

(IMC 2014)