, 2022

Since $3m^3$ is a multiple of
3, then $5n^5$ is a multiple of
3.

Since $5$ is not a multiple of 3 and
3 is a prime number, then $n^5$ is a
multiple of 3.

Since $n^5$ is a multiple of 3 and 3
is a prime number, then $n$ is a
multiple of 3, which means that $5n^5$ includes at least 5 factors of
3.

Since $5n^5$ includes at least 5
factors of 3, then $3m^3$ includes at
least 5 factors of 3, which means that $m^3$ is a multiple of 3, which means that
$m$ is a multiple of 3.

Using a similar analysis, both $m$
and $n$ must be multiples of 5.

Therefore, we can write $m = 3^a 5^b
s for some positive integers a$,$b$
and $s$ and we can write $n = 3^c 5^d t$ for some positive integers
$c$, $d$ and $t$, where neither $s$ nor $t$ is a multiple of 3 or 5. (In other
words, we have grouped all of the factors of 3 and 5 in each of $m$ and $n$.)

From the given equation, $$\begin{align*} 3m^3 & = 5n^5 \\ 3(3^a 5^b s)^3 & = 5(3^c 5^d t)^5 \\ 3 \times 3^{3a} 5^{3b} s^3 & = 5 \times 3^{5c} 5^{5d} t^5 \\ 3^{3a+1} 5^{3b} s^3 & = 3^{5c} 5^{5d+1} t^5\end{align*}$$
Since $s$ and $t$ are not multiples of 3 or 5, we must
have $3^{3a+1} = 3^{5c}$ and $5^{3b} = 5^{5d+1}$ and $s^3 = t^5$.

Since $s$ and $t$ are positive and $m$ and $n$ are to be as small as possible, we can
set $s=t=1$, which satisfy $s^3 = t^5$.

Since $3^{3a+1} = 3^{5c}$ and $5^{3b} = 5^{5d+1}$, then $3a+1 = 5c$ and $3b = 5d+1$.

Since $m$ and $n$ are to be as small as possible, we want
to find the smallest positive integers $a,b,c,d$ for which $3a+1=5c$ and $3b=5d+1$.

Neither $a=1$ nor $a=2$ gives a value for $3a+1$ that is a multiple of 5, but $a=3$ gives $c=2$.

Similarly, $b=1$ does not give a
value of $3b$ that equals $5d+1$ for any positive integer $d$, but $b=2$ gives $d=1$.

Therefore, the smallest possible values of $m$ and $n$ are $m = 3^3
5^2 = 675$and$n = 3^2 5^1 =
45$, which gives$m+n =
720$.

(We can verify by substitution that $m =
675$and$n = 45$ satisfy the
equation $3m^3 = 5n^5$.)