2003 AMC 10B Problem 16

P16-20~2 minPrint

Basic Counting Absolute Value & Inequalities Bound the quantity above/below or estimate to pin it down Set up the equation/formula and compute; no special trick needed

A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that a restaurant should offer so that a customer could have a different dinner each night in the year $2003$ ?

Show solution without answering

Key insight

There are 6 times m times m dinners, which must reach 365. Seven main courses give 294, too few; eight give 384.

Solution

1. Call the number of main courses $m$ . Then there are $2m$ appetizers and $3$ desserts.
2. A dinner picks one of each, so the number of different dinners is
$2m \times m \times 3 = 6m^2.$
3. The year $2003$ is not a leap year ( $2003$ is not divisible by $4$ ), so it has $365$ days. We need at least $365$ different dinners.
4. Try the choices, starting from the smallest:

$m$	$6m^2$	Enough for $365$ ?
$4$	$96$	no
$5$	$150$	no
$6$	$216$	no
$7$	$294$	no
$8$	$384$	yes

So $7$ main courses fall short and $8$ are enough.

The answer is $\boxed{\textbf{(E)}\ 8}$ .

Why this works

Independent choices multiply, so the number of dinners grows like $6m^2$ . "Enough for every day of the year" means $6m^2 \ge 365$ , and the answer is the smallest whole number $m$ that makes this true. Always check both sides: the answer works and the number just below it does not.

Concise solution

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

With $m$ main courses there are $2m$ appetizers and $3$ desserts, so by the multiplication principle the number of distinct dinners is
$2m \cdot m \cdot 3 = 6m^2.$

The year $2003$ is not a leap year ( $2003$ is not divisible by $4$ ), so it has $365$ days, and we need $6m^2 \ge 365$ , i.e. $m^2 \ge 60.8\ldots$ .

Since $7^2 = 49 < 61$ and $8^2 = 64 \ge 61$ , the smallest possible $m$ is $8$ . Check: $6 \cdot 64 = 384 \ge 365$ , while $6 \cdot 49 = 294 < 365$ .

The answer is $\boxed{\textbf{(E)}\ 8}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)