2003 AMC 10B Problem 10

P1-10~1 minPrint

Basic Counting Exponents, Logarithms & Radicals Set up the equation/formula and compute; no special trick needed

Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times has the number of possible license plates increased?

Show solution without answering

Key insight

Old plates: 26 times 10 four times. New plates: 26 three times times 10 three times. Divide new by old and cancel: 26 squared over 10.

Solution

1. Each letter spot can be any of $26$ letters, and each digit spot any of $10$ digits. The spots are filled independently, so multiply the choices for each spot.
2. Old plate (letter, digit, digit, digit, digit): $26 \times 10 \times 10 \times 10 \times 10 = 26 \cdot 10^4$ .
3. New plate (letter, letter, letter, digit, digit, digit): $26 \times 26 \times 26 \times 10 \times 10 \times 10 = 26^3 \cdot 10^3$ .
4. "How many times" means divide new by old:
$\frac{26^3 \cdot 10^3}{26 \cdot 10^4}.$
One $26$ on the bottom cancels one of the three on top, leaving $26^2$ . Three $10$ s on top cancel three of the four on the bottom, leaving one $10$ on the bottom. The result is $\frac{26^2}{10}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{26^2}{10}}$ .

Why this works

When choices are made one after another and do not affect each other, the number of results is the product of the number of options at each step. Comparing two such products "by how many times" is a division, and matching factors cancel. Writing out which factors cancel keeps the powers straight.

Concise solution

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

Each letter position can be filled in $26$ ways and each digit position in $10$ ways, independently, so the number of plates is the product over positions.

Old scheme (one letter, four digits): $26 \cdot 10^4$ .

New scheme (three letters, three digits): $26^3 \cdot 10^3$ .

"By how many times" asks for the ratio new to old:
$\frac{26^3 \cdot 10^3}{26 \cdot 10^4} = \frac{26^2}{10}.$

The answer is $\boxed{\textbf{(C)}\ \frac{26^2}{10}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)