AMC 8 · 2008 · #22

Grade 6 number-theory

Archetype Small-Domain Constrained Search

divisibility-rulesmultiplesinterval-arithmeticsequences-arithmetic bound-inequality-then-enumerateidentify-subproblems ↑ Prerequisites: divisibility-rulesmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: For how many positive integers $n$ are both $\dfrac{n}{3}$ and $3n$ three-digit whole numbers?

Givens: $n$ is a positive integer; $\dfrac{n}{3}$ must be a three-digit whole number (so it is an integer from $100$ to $999$); $3n$ must be a three-digit whole number (so it is an integer from $100$ to $999$); Answer choices: (A) $12$, (B) $21$, (C) $27$, (D) $33$, (E) $34$

Unknowns: The count of positive integers $n$ that satisfy both conditions

Understand

Restated: For how many positive integers $n$ are both $\dfrac{n}{3}$ and $3n$ three-digit whole numbers?

Plan

Primary tool: #7 Break Into Subproblems

Secondary: #13 Count Carefully

The problem stacks two separate conditions onto the same $n$, so Tool #7 (Break Into Subproblems) says: handle each condition by itself, then combine. Condition 1 ($\tfrac{n}{3}$ is a three-digit whole number) pins down a range for $n$ and forces $n$ to be a multiple of $3$. Condition 2 ($3n$ is a three-digit whole number) pins down another range for $n$. Intersecting the two ranges leaves a short list, and Tool #13 (Count Carefully) gives the final count of multiples of $3$ in that overlap.

Execute — Answer: A

#7 Break Into Subproblems 6.EE.B.8 Step 1

Turn the first condition into an inequality on $n$.
$\dfrac{n}{3}$ is a three-digit whole number means $100 \le \dfrac{n}{3} \le 999$.
Multiply every part by $3$.

$$100 \le \dfrac{n}{3} \le 999 \;\Rightarrow\; 300 \le n \le 2997$$

💡 Writing the size constraint as an inequality is the Grade 6 "write an inequality for a real-world constraint" move.

#7 Break Into Subproblems 4.OA.B.4 Step 2

The same condition also says $\dfrac{n}{3}$ is a whole number, so $n$ must be a multiple of $3$.
Keep this divisibility rule alongside the range.

$n = 3k$ for some positive integer $k$

💡 Grade 4 multiples: $\tfrac{n}{3}$ is a whole number exactly when $n$ is in the list $3, 6, 9, 12, \ldots$

#7 Break Into Subproblems 6.EE.B.8 Step 3

Turn the second condition into an inequality on $n$.
$3n$ is a three-digit whole number means $100 \le 3n \le 999$.
Divide every part by $3$.

$$100 \le 3n \le 999 \;\Rightarrow\; \dfrac{100}{3} \le n \le 333 \;\Rightarrow\; 34 \le n \le 333$$

💡 Since $n$ is an integer and $\tfrac{100}{3} \approx 33.3$, the smallest integer that works is $34$. ($3n$ is automatically a whole number when $n$ is.)

#7 Break Into Subproblems 6.EE.B.5 Step 4

Intersect the two ranges.
$n$ must satisfy $300 \le n \le 2997$ and $34 \le n \le 333$ at the same time, so $n$ lies in the overlap.

$$300 \le n \le 333$$

💡 Grade 6 inequality reasoning: keep only the values that satisfy every condition.

#13 Count Carefully 4.OA.C.5 Step 5

Count the multiples of $3$ in $[300, 333]$.
The list is $300, 303, 306, \ldots, 333$.
Use the arithmetic-sequence count: $\dfrac{\text{last} - \text{first}}{\text{step}} + 1$.

$$\dfrac{333 - 300}{3} + 1 = \dfrac{33}{3} + 1 = 11 + 1 = 12 \;\Rightarrow\; \textbf{(A)}$$

💡 Counting evenly-spaced terms is the Grade 4 "generate and analyze a pattern" idea: $12$ multiples of $3$ fit in this stretch.

[1] #7 6.EE.B.8 Turn the first condition into an inequality on $n$. $\dfrac{n}{3}$ is a three-di

[2] #7 4.OA.B.4 The same condition also says $\dfrac{n}{3}$ is a whole number, so $n$ must be a

[3] #7 6.EE.B.8 Turn the second condition into an inequality on $n$. $3n$ is a three-digit whole

[4] #7 6.EE.B.5 Intersect the two ranges. $n$ must satisfy $300 \le n \le 2997$ and $34 \le n \l

[5] #13 4.OA.C.5 Count the multiples of $3$ in $[300, 333]$. The list is $300, 303, 306, \ldots,

Review

Reasonableness: Check the endpoints. At $n = 300$: $\tfrac{n}{3} = 100$ (smallest three-digit number) and $3n = 900$ (still three-digit). At $n = 333$: $\tfrac{n}{3} = 111$ (still three-digit) and $3n = 999$ (largest three-digit number). Both ends pass, so the window $[300, 333]$ is exactly right, and the $12$ multiples of $3$ inside it are all valid.

Alternative: Tool #4 (Introduce a Variable): let $x = \tfrac{n}{3}$, so $n = 3x$ and $3n = 9x$. The conditions become "$x$ is a three-digit whole number" and "$9x$ is a three-digit whole number". The first gives $100 \le x \le 999$; the second gives $100 \le 9x \le 999$, i.e. $12 \le x \le 111$. Intersect: $100 \le x \le 111$, which holds for $111 - 100 + 1 = 12$ integers $x$, matching answer (A).

CCSS standards used (min grade 6)

4.OA.B.4 Find all factor pairs for a whole number; recognize multiples (Translating "$\tfrac{n}{3}$ is a whole number" into "$n$ is a multiple of $3$".)
4.OA.C.5 Generate a number or shape pattern that follows a given rule (Listing $300, 303, 306, \ldots, 333$ as the multiples of $3$ in the final window and using the arithmetic-sequence count.)
6.EE.B.8 Write an inequality of the form $x > c$ or $x < c$ to represent a constraint (Writing $100 \le \tfrac{n}{3} \le 999$ and $100 \le 3n \le 999$ from the three-digit conditions.)
6.EE.B.5 Understand solving an inequality as a process of answering which values make it true (Intersecting the two ranges to get $300 \le n \le 333$ as the set of $n$ that satisfy both conditions.)

⭐ Two rules on the same $n$? Turn each one into its own range, then keep only the $n$ values that fit both — and remember the divisibility rule from the fraction. After that, this AMC 8 problem is just counting multiples of $3$ from $300$ to $333$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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