12. Which consecutive positive integers sum to $6^{3}$? Find all solutions.
12. Solution: Let there be $n$ consecutive positive integers $(n>1)$. The smallest one is $a$, then $a+(a+1)+\cdots+(a+n-1)=6^{3}$.
That is, $n(2a+n-1)=432$.
Obviously, $n \neq 2a+n-1(\bmod 2)$, which means $n$ and $2a+n-1$ have different parity.
Since $a \geqslant 1$, it follows that $2a+n-1>n$.
Given $432=2^{4} \cdot 3^{3}, n<\sqrt{432}<21$,
the possible values for $n$ are $2,3,4,6,8,9,12,16,18$.
Only when $n$ is $3,9,16$, $\frac{432}{n}$ and $n$ have different parity.
Therefore, there are only three solutions: $(n, a)=(3,71),(9,20),(16,6)$.