If $m$ and $n$ are positive integers that satisfy the equation $3 m^{3}=5 n^{5}$, the smallest possible value for $m+n$ is
(A) 900
(B) 450
(C) 720
(D) 810
(E) 630

Part C: Each correct answer is worth 8.

Each correct answer is an integer from 0 to 99, inclusive.

A one-digit answer (such as " 7 ") must be coded with a leading zero (" 07 ").

Note: The integer formed by the rightmost two digits of 12345 is 45 . The integer formed by the rightmost two digits of 6307 is 7 , coded 07 .
Since $3 m^{3}$ is a multiple of 3 , then $5 n^{5}$ 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 $5 n^{5}$ includes at least 5 factors of 3 .

Since $5 n^{5}$ includes at least 5 factors of 3 , then $3 m^{3}$ 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,

$$
\begin{aligned}
3 m^{3} & =5 n^{5} \\
3\left(3^{a} 5^{b} s\right)^{3} & =5\left(3^{c} 5^{d} t\right)^{5} \\
3 \times 3^{3 a} 5^{3 b} s^{3} & =5 \times 3^{5 c} 5^{5 d} t^{5} \\
3^{3 a+1} 5^{3 b} s^{3} & =3^{5 c} 5^{5 d+1} t^{5}
\end{aligned}
$$

Since $s$ and $t$ are not multiples of 3 or 5 , we must have $3^{3 a+1}=3^{5 c}$ and $5^{3 b}=5^{5 d+1}$ 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 $3^{3 a+1}=3^{5 c}$ and $5^{3 b}=5^{5 d+1}$, then $3 a+1=5 c$ and $3 b=5 d+1$.

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 $3 a+1=5 c$ and $3 b=5 d+1$.

Neither $a=1$ nor $a=2$ gives a value for $3 a+1$ that is a multiple of 5 , but $a=3$ gives $c=2$. Similarly, $b=1$ does not give a value of $3 b$ that equals $5 d+1$ 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 $3 m^{3}=5 n^{5}$.)

ANSWER: (C)