AMC 10 · 2003 · #16

Grade 6 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

fundamental-counting-principlelinear-equations-one-var bound-inequality-then-enumerateguess-and-check ↑ Prerequisites: fundamental-counting-principle

📏 Medium solution 💡 2 insights

Problem

A restaurant offers three desserts and exactly twice as many appetizers as main courses. A dinner is one appetizer, one main course, and one dessert. What is the least number of main courses the restaurant must offer so that a customer can have a different dinner every night of the year 2003?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The three menu sizes are all tied together, so tool #4 (Introduce a Variable) is the fastest way in: name the number of main courses m, then the appetizers are 2m and the desserts stay 3. Because you build a dinner by picking one from each list independently, the number of different dinners is the product 2m · m · 3 = 6m². The question becomes: what is the smallest whole m with 6m² ≥ 365? Tool #6 (Guess and Check) finishes it cleanly — the answers are small whole numbers, so testing m = 7 and m = 8 against 365 settles it, and tool #3 (Eliminate Possibilities) uses the fact that dinners only grow as m grows to rule out everything below the first value that works.

1STEP 1

Name the main courses m

Let m be the number of main courses. Then the appetizers number 2m, and the desserts stay fixed at 3.

main courses = m, appetizers = 2m, desserts = 3

2STEP 2

Count the dinners as a product

Independent choices multiply, so the dinner count is 2m · m · 3 = 6m².

dinners = (2m)(m)(3) = 6m²

3STEP 3

Set up the day count

2003 is not divisible by 4, so it has 365 days, and the dinners must reach that: 6m² ≥ 365.

6m² ≥ 365

4STEP 4

Test the candidates

6m² only grows, so step up: m = 7 gives 294, short of 365; m = 8 gives 384. Answer (E).

6 · 7² = 294 ＜ 365, 6 · 8² = 384 ≥ 365 → (E)

Answer

Check the boundary directly: with m = 8 there are 16 appetizers, 8 main courses, and 3 desserts, giving 16 · 8 · 3 = 384 dinners — comfortably past the 365 days needed, with 19 to spare. With m = 7 there are only 14 · 7 · 3 = 294 dinners, short of 365 by 71. So 8 is exactly the tipping point, matching (E). Since 6m² jumps in bigger steps as m climbs, no value between 7 and 8 exists to try.

💡Key takeaway

Name the unknown, multiply the menu sizes to count the meals, then step the number up until you have enough for all 365 days.

Name the main courses m
Count the dinners as a product
Set up the day count
Test the candidates

Name the main courses m

Count the dinners as a product

Set up the day count

Test the candidates

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