2003 AMC 10B Problem 14

1. We want to write $3^8 \cdot 5^2$ as some whole number $a$ raised to a power $b$. A bigger $b$ makes $a$ much smaller, so try to make $b$ as big as possible.
2. If $a$ is multiplied by itself $b$ times, every prime inside $a$ shows up $b$ times as often. So the exponent $8$ on the $3$ and the exponent $2$ on the $5$ must both be multiples of $b$.
3. The numbers that go evenly into both $8$ and $2$ are $1$ and $2$. So $b$ is $1$ or $2$.
4. Try both:
- $b = 1$: $a = 3^8 \cdot 5^2 = 164025$, and $a + b = 164026$.
- $b = 2$: $a = 3^4 \cdot 5 = 81 \cdot 5 = 405$, because $405 \cdot 405 = 3^8 \cdot 5^2$. Then $a + b = 407$.
5. The smaller total is $407$.

The answer is $\boxed{\textbf{(D)}\ 407}$.

A number is a perfect square, cube, and so on exactly when every exponent in its prime factorization is a multiple of that power. So the possible $b$ are the numbers that divide all the exponents. Here that is at most $2$, and the largest allowed $b$ gives the smallest $a$.

Use the choices. Since $b$ is at most $2$, $a$ is at least $405$ (the square root of $3^8 \cdot 5^2$). So (A), (B) and (C) are too small. And $405 + 2 = 407$ is smaller than $900$.

Write $a$ in terms of its prime factorization. Since $a^b = 3^8 \cdot 5^2$, the primes of $a$ are only $3$ and $5$, say $a = 3^p 5^q$, and then $bp = 8$ and $bq = 2$. So $b$ divides both $8$ and $2$, which forces $b \in \{1, 2\}$.

• $b = 1$: $a = 3^8 \cdot 5^2 = 164025$, so $a + b = 164026$.
• $b = 2$: $a = 3^4 \cdot 5 = 81 \cdot 5 = 405$, since $405^2 = 3^8 \cdot 5^2$. Then $a + b = 407$.

The smaller total is $407$.

The answer is $\boxed{\textbf{(D)}\ 407}$.