AMC 8 · 2008 · #22
Grade 6 number-theoryProblem
For how many positive integer values of n are both 3n and 3n three-digit whole numbers?
Pick an answer.
AMC 8 2008 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.
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?
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) $34$
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
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 $3$.
💡 Writing the size constraint as an inequality is the Grade 6 "write an inequality for a real-world constraint" move.
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.
💡 Grade 4 multiples: $\tfrac{n}{3}$ is a whole number exactly when $n$ is in the list $3, 6, 9, 12, \ldots$
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 $3$.
💡 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.)
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.
💡 Grade 6 inequality reasoning: keep only the values that satisfy every condition.
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: $\dfrac{\text{last} - \text{first}}{\text{step}} + 1$.
💡 Counting evenly-spaced terms is the Grade 4 "generate and analyze a pattern" idea: $12$ multiples of $3$ fit in this stretch.
6.EE.B.8 Turn the first condition into an inequality on $n$. $\dfrac{n}{3}$ is a three-di 4.OA.B.4 The same condition also says $\dfrac{n}{3}$ is a whole number, so $n$ must be a 6.EE.B.8 Turn the second condition into an inequality on $n$. $3n$ is a three-digit whole 6.EE.B.5 Intersect the two ranges. $n$ must satisfy $300 \le n \le 2997$ and $34 \le n \l 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.4Find 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.5Generate 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.8Write 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.5Understand 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 $333$.
⭐ 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 $333$.
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