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2003 AMC 10B Problem 16

2003 AMC 10B Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 2 of 5 · about 2 min · position P16-20

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 $2003$ ?

A)
$4$
B)
$5$
C)
$6$
D)
$7$
E)
$8$

Answer

Key insight

There are 2m * m * 3 = 6m^2 dinners and 365 days, so m^2 >= 61 and the least m is 8.

Solution

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 $\boxed{\textbf{(E)}\ 8}$ .

Why this works

Independent choices multiply, giving a count that is a quadratic in $m$ ; "enough for a year" is an inequality, and the answer is the least integer satisfying it. Always verify the boundary on both sides: the candidate works and the one below it fails.

The trap

Rounding m = sqrt(60.8) down to 7 even though 6 * 49 = 294 falls short of 365.

Common mistakes

Rounding $m = \sqrt{60.8}$ down to $7$ even though $6 \cdot 49 = 294$ falls short of $365$ .
Forgetting the factor $3$ for desserts (solving $2m^2 \ge 365$ , giving $14$ ) or using $366$ days, which happens not to change the answer.

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed