5. (5 points) The decimal counting method is to carry 1 when reaching 10, such as $24_{10}=2 \times 10+4 \times 1,36510=3 \times 10^{2}+6 \times 10+5 \times 1$; computers use the binary counting method, which is to carry 1 when reaching 2, such as $710=1 \times 2^{2}+1 \times 2+1 \times 1=111_{2}$, $12_{10}=1 \times 2^{3}+1 \times 2^{2}+0 \times 2+0 \times 1=1100$, if a natural number can be written as the $m$-ary number $45 m$, it can also be written as the $n$-ary number $54_{n}$, then the smallest $m=$ $\qquad$ , $n=$ $\qquad$ . (Note:
$$
a^{n}=\underbrace{a \times a \times a \times \cdots \times a}_{n \uparrow a})
$$
【Solution】Solution: $45_{m}=4 m+5$;
$$
54_{n}=5 n+4 ;
$$

Then:
$$
4 m+5=5 n+4
$$

That is: $4(m-1)=5(n-1)$,
If $m-1=5, n-1=4$, then $m=6, n=5$, but in this case, the digit 5 cannot appear in base $n$; if $m-1=10, n-1=8$, then $m=11, n=9$, which meets the requirement. That is, the smallest value of $m$ is $11$, and the smallest value of $n$ is $9$. Therefore, the answer is: 11,9.