2003 AMC 10B Problem 14

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 $\boxed{\textbf{(D)}\ 407}$.

A number is a perfect $b$-th power exactly when every exponent in its prime factorization is a multiple of $b$, so the possible exponents are the common divisors of those exponents. Here $\gcd(8, 2) = 2$ caps $b$ at $2$, and a larger $b$ always shrinks $a$ dramatically, so the largest legal $b$ minimizes $a + b$.

Answer-choice sanity check: since $b \le 2$, $a$ is at least $\sqrt{3^8 \cdot 5^2} = 405$, ruling out (A), (B), (C) immediately; $405 + 2 = 407$ beats $900$.

• Trying to make $b$ as large as possible, e.g. $b = 8$ with a non-integer $a$, or taking $b = 1$ and adding the huge base.
• Arithmetic slip in $3^4 \cdot 5$ (e.g. $81 \cdot 5 = 305$), which makes the sum $307$ and matches nothing.