2003 AMC 10B Problem 14

P11-15~2 minPrint

Exponents, Logarithms & Radicals Number Properties Set up the equation/formula and compute; no special trick needed Consider the largest/smallest element or boundary case

Given that $3^8\cdot5^2=a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b$ .

Show solution without answering

Key insight

The exponent b must go evenly into both 8 and 2, so b is at most 2. Then a = 405 and a + b = 407.

Solution

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

Why this works

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 $a$ .

Alternative approach

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 $900$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)
Introduction to Algebra, Chapter 20: Special Functions (§20.3)