AMC 8 · 2008 · #22

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

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

📏 Medium solution 💡 3 insights

Problem

For how many positive integers n are both

\frac{n}{3}

and 3n three-digit whole numbers?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Break Into Subproblems

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 ( $\frac{n}{3}$ 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.

1STEP 1

Turn condition 1 into a range: 100 ≤ $\frac{n}{3}$ ≤ 999, then multiply each part by 3 to get 300 ≤ n ≤ 2997.

100 ≤

\frac{n}{3}

≤ 999 → 300 ≤ n ≤ 2997

2STEP 2

The same condition also makes $\frac{n}{3}$ a whole number, so n must be a multiple of 3.

n = 3k for some positive integer k

3STEP 3

Turn condition 2 into a range: 100 ≤ 3n ≤ 999, then divide each part by 3 to get 34 ≤ n ≤ 333.

100 ≤ 3n ≤ 999 →

\frac{100}{3}

≤ n ≤ 333 → 34 ≤ n ≤ 333

4STEP 4

Intersect the two ranges: 300 ≤ n ≤ 2997 and 34 ≤ n ≤ 333 overlap in 300 ≤ n ≤ 333.

300 ≤ n ≤ 333

5STEP 5

Count the multiples of 3 in [300, 333]: (333 - 300)/3 + 1 = 12.

\frac{333 - 300}{3}

+ 1 =

\frac{33}{3}

+ 1 = 11 + 1 = 12 → (A)

Answer

Check the endpoints. At n = 300:

\frac{n}{3}

= 100 (smallest three-digit number) and 3n = 900 (still three-digit). At n = 333:

\frac{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.

💡Key takeaway

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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