Concatenation with All the Digits

Note that $n^3$ and $n^4$ must total ten digits between them. $17^3$ has four digits and $17^4$ has five digits, so $n>17$ . Similarly, $22^3$ has five digits and $22^4$ has six digits, so $n<22$ . Furthermore, $n$ cannot end in a $0$ or $1$ , since this would lead to $n^3$ and $n^4$ both ending in $0$ and $1$ , respectively. This leaves $18$ and $19$ as possibilities: $\begin{array}{c|c|c} n & n^3 & n^4 \\ \hline 18 & 5832 & 104976 \\ 19 & 6859 & 130321 \end{array}$ Thus $n=\boxed{18}$ .