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 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 $2003$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A restaurant has $3$ desserts and twice as many appetizers as main courses. A dinner is one appetizer, one main course, and one dessert. Find the smallest number of main courses so the restaurant can serve a different dinner on every day of the year $2003$.

Givens: There are exactly $3$ desserts.; The number of appetizers is twice the number of main courses.; A dinner is one appetizer AND one main course AND one dessert, chosen independently.; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$

Unknowns: The least number of main courses the restaurant must offer

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #6 Guess and Check, #3 Eliminate Possibilities

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 $2m \cdot m \cdot 3 = 6m^2$. The question becomes: what is the smallest whole $m$ with $6m^2 \ge 365$? 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.

Execute — Answer: E

#4 Introduce a Variable 6.EE.B.6 Step 1

Name the main courses m

Let $m$ be the number of main courses.
The problem says there are twice as many appetizers as main courses, so the number of appetizers is $2m$.
The number of desserts is fixed at $3$.
Now every menu size is written in terms of the one unknown $m$.

$$\text{main courses} = m, \quad \text{appetizers} = 2m, \quad \text{desserts} = 3$$

💡 Give the one thing you don't know a name, and everything else on the menu is measured against it.

#4 Introduce a Variable 3.OA.A.1 Step 2

Count the dinners as a product

A dinner is one appetizer, then one main course, then one dessert, each chosen on its own.
When choices are made independently one after another, you multiply the number of options at each step.
So the number of different dinners is $2m$ (appetizers) times $m$ (main courses) times $3$ (desserts).
Multiplying, $2m \cdot m \cdot 3 = 6m^2$, since $m \cdot m = m^2$.

$$\text{dinners} = (2m)(m)(3) = 6m^2$$

💡 Pairing every appetizer with every main course with every dessert multiplies the three list lengths together.

#4 Introduce a Variable 6.EE.B.8 Step 3

Set up the day count

The customer wants a different dinner every night of $2003$.
Since $2003$ is not a leap year ($2003$ is not divisible by $4$), it has $365$ days.
So the number of dinners must be at least $365$.
Setting the dinner count against the days gives the requirement $6m^2 \ge 365$.

$$6m^2 \ge 365$$

💡 Enough dinners for the whole year means the dinner count must reach the number of days, $365$.

#6 Guess and Check 6.EE.B.5 Step 4

Test the candidates

The dinner count $6m^2$ only grows as $m$ grows, so find the first whole $m$ that reaches $365$.
Try $m = 7$: $6 \cdot 7^2 = 6 \cdot 49 = 294$, which is less than $365$ — not enough.
Try $m = 8$: $6 \cdot 8^2 = 6 \cdot 64 = 384$, which is at least $365$ — enough.
So $7$ main courses fall short and $8$ is the smallest number that works.
The answer is (E) $8$.

$$6\cdot 7^2 = 294 < 365, \qquad 6\cdot 8^2 = 384 \ge 365 \;\Rightarrow\; \textbf{(E)}$$

💡 Walk $m$ up one step at a time; the first value whose dinner count clears $365$ is the answer.

[1] #4 6.EE.B.6 Let $m$ be the number of main courses. The problem says there are twice as many

[2] #4 3.OA.A.1 A dinner is one appetizer, then one main course, then one dessert, each chosen o

[3] #4 6.EE.B.8 The customer wants a different dinner every night of $2003$. Since $2003$ is not

[4] #6 6.EE.B.5 The dinner count $6m^2$ only grows as $m$ grows, so find the first whole $m$ tha

Review

Reasonableness: Check the boundary directly: with $m = 8$ there are $16$ appetizers, $8$ main courses, and $3$ desserts, giving $16 \cdot 8 \cdot 3 = 384$ dinners — comfortably past the $365$ days needed, with $19$ to spare. With $m = 7$ there are only $14 \cdot 7 \cdot 3 = 294$ dinners, short of $365$ by $71$. So $8$ is exactly the tipping point, matching (E). Since $6m^2$ jumps in bigger steps as $m$ climbs, no value between $7$ and $8$ exists to try.

Alternative: Solve the inequality instead of testing: $6m^2 \ge 365$ gives $m^2 \ge 60.8\ldots$, so $m \ge \sqrt{60.8\ldots} \approx 7.8$. The smallest whole number at least $7.8$ is $8$, again landing on (E).

CCSS standards used (min grade 6)

6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Letting $m$ stand for the main courses and writing the appetizers as $2m$.)
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups (Multiplying the three menu sizes to count the dinners as $2m \cdot m \cdot 3 = 6m^2$.)
6.EE.B.8 Write an inequality of the form x > c or x < c to represent a constraint (Turning "a different dinner every day of $2003$" into $6m^2 \ge 365$ using the $365$ non-leap-year days.)
6.EE.B.5 Understand solving an inequality as a process of finding values that make it true (Testing $m = 7$ and $m = 8$ to find the least whole number with $6m^2 \ge 365$.)

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

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Name the main courses m

Count the dinners as a product

Set up the day count

Test the candidates

Review

CCSS standards used (min grade 6)

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