Is there a systematic way to find a permutation with a power equal to this product of transpositions? I am trying to find an n-cycle, $\sigma$, $n\ge 10$ with 
$$\sigma^k=(12)(34)(56)(78)(9\; 10)$$
For k an integer in a systematic way. 
I know the basic facts about powers of $n$ cycles, as in 
$$
\sigma^n=e\\
$$
and $\sigma^{-1}$ just reflects the cycle horizontally. Can I use these?
 A: Powering a permutation splits its cycles into smaller ones all of the same length. In particular, this means that $\sigma$ must be a permutation of $\{1,\cdots,10\}$ times a permutation of all numbers from $11$ to $n$. The latter must be something of order $k$ in order to disappear after powering. Without loss of generality, assume $\sigma$ is simply a permutation of $\{1,\cdots,10\}$.
No odd-length cycle can appear in $\sigma$ since the entries of the odd-length cycle would never appear in an even-length cycle of $\sigma^k$, but $\sigma^k$'s even-length cycles include all numbers $1,\cdots,10$.
Thus the cycle type of $\sigma$ must be among the list


*

*$10$

*$8,2$

*$6,4$

*$6,2,2$

*$4,4,2$

*$4,2,2,2$

*$2,2,2,2,2$


If $k$ is odd it will preserve the cycle type, so it must be $2,2,2,2,2$. Then the only solution is
$$ \sigma=(\color{DarkOrange}{12})(\color{Red}{34})(\color{Purple}{56})(\color{Blue}{78})(\color{Green}{9\,10}). $$
If $k$ is even, then powering kills $2$-cycles, leaving only types $10$ and $6,4$ as possibilities.
Type $10$ powers to type $2,2,2,2,2$ precisely if $k$ is divisible by $5$ but not $2$. One solution is
$$ \sigma=(\color{DarkOrange}{1}~\color{Red}{3}~\color{Purple}{5}~\color{Blue}{7}~\color{Green}{9}~\color{DarkOrange}{2}~\color{Red}{4}~\color{Purple}{6}~\color{Blue}{8}~\color{Green}{10}). $$
Type $6,4$ powers to type $2,2,2,2,2$ precisely if $k$ is divisible by $2$ and $3$ but not $4$, and
$$ \sigma=(\color{DarkOrange}{1}~\color{Red}{3}~\color{Purple}{5}~\color{DarkOrange}{2}~\color{Red}{4}~\color{Purple}{6})(\color{Blue}{7}~\color{Green}{9}~\color{Blue}{8}~\color{Green}{10}) $$
is a solution.
I've colored the solutions in order to illustrate the interlacing process I used to create them. In order to obtain every possible solution, simply rearrange $(12)(34)(56)(78)(9\,10)$ by permuting the cycles and transposing their entries, then interlace as above. (Apologies to the color-blind.)
