AMC 8 · 2003 · #19

Grade 6 number-theory

Archetype Arithmetic Expression Decomposition

lcmmultiplesprime-factorizationdivisibility-rules identify-subproblemssystematic-enumeration ↑ Prerequisites: prime-factorizationmultiplesmulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Count the integers strictly between $1000$ and $2000$ that are divisible by all three of $15, 20,$ and $25$.

Givens: The integer $N$ must satisfy $1000 < N < 2000$; $15 \mid N$, $20 \mid N$, and $25 \mid N$ all at once; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: The count of integers $N$ that meet every condition

Understand

Restated: Count the integers strictly between $1000$ and $2000$ that are divisible by all three of $15, 20,$ and $25$.

Givens: The integer $N$ must satisfy $1000 < N < 2000$; $15 \mid N$, $20 \mid N$, and $25 \mid N$ all at once; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Plan

Primary tool: #7 Break into Subproblems

Secondary: #2 Make a Systematic List

Three divisibility rules at once feels like three problems, but Tool #7 (Break into Subproblems) splits it cleanly: first collapse the three rules into one by finding $\operatorname{lcm}(15, 20, 25)$; then count how many multiples of that LCM land in the open interval $(1000, 2000)$. Once the LCM is in hand, Tool #2 (Make a Systematic List) finishes the job — there are only a few candidates, so listing them is faster and safer than algebra.

Execute — Answer: C

#7 Break into Subproblems 6.NS.B.4 Step 1

Subproblem 1: turn the three divisibility conditions into one.
A number divisible by $15, 20,$ and $25$ is a common multiple of all three, and the common multiples are exactly the multiples of $\operatorname{lcm}(15, 20, 25)$.

$$15 \mid N \text{ and } 20 \mid N \text{ and } 25 \mid N \;\Longleftrightarrow\; \operatorname{lcm}(15, 20, 25) \mid N$$

💡 One LCM rule replaces three separate divisibility checks. That is the Grade 6 "least common multiple" idea in action.

#7 Break into Subproblems 6.NS.B.4 Step 2

Compute the LCM from prime factorizations.
Write each number as a product of primes, then take the highest power of each prime that appears anywhere.

$$15 = 3 \cdot 5, \;\; 20 = 2^2 \cdot 5, \;\; 25 = 5^2 \;\Rightarrow\; \operatorname{lcm} = 2^2 \cdot 3 \cdot 5^2 = 4 \cdot 3 \cdot 25 = 300$$

💡 Highest power of $2$ is $2^2$ (from $20$); of $3$ is $3^1$ (from $15$); of $5$ is $5^2$ (from $25$). Multiply: $4 \cdot 3 \cdot 25 = 300$.

#2 Make a Systematic List 4.OA.B.4 Step 3

Subproblem 2: list the multiples of $300$ that fall strictly between $1000$ and $2000$.
Start at the first multiple of $300$ above $1000$ and walk upward by $300$ until passing $2000$.

$$300 \cdot 4 = 1200, \;\; 300 \cdot 5 = 1500, \;\; 300 \cdot 6 = 1800, \;\; 300 \cdot 7 = 2100$$

💡 $300 \cdot 3 = 900$ is below the range and $300 \cdot 7 = 2100$ is above it. Only the values strictly between $1000$ and $2000$ count.

#2 Make a Systematic List 4.OA.A.3 Step 4

Count the valid multiples.
The numbers $1200, 1500, 1800$ all lie inside the open interval, while $900$ and $2100$ do not.

$$\{1200, \; 1500, \; 1800\} \;\Rightarrow\; 3 \text{ integers} \;\Rightarrow\; \textbf{(C)}$$

💡 A short systematic list makes the count obvious: three multiples fit.

[1] #7 6.NS.B.4 Subproblem 1: turn the three divisibility conditions into one. A number divisibl

[2] #7 6.NS.B.4 Compute the LCM from prime factorizations. Write each number as a product of pri

[3] #2 4.OA.B.4 Subproblem 2: list the multiples of $300$ that fall strictly between $1000$ and

[4] #2 4.OA.A.3 Count the valid multiples. The numbers $1200, 1500, 1800$ all lie inside the ope

Review

Reasonableness: Spot-check that each listed number really is divisible by $15, 20, 25$: $1200 = 15 \cdot 80 = 20 \cdot 60 = 25 \cdot 48$; $1500 = 15 \cdot 100 = 20 \cdot 75 = 25 \cdot 60$; $1800 = 15 \cdot 120 = 20 \cdot 90 = 25 \cdot 72$. All three pass. The boundary multiples $900$ and $2100$ fall outside $(1000, 2000)$, so they are correctly excluded. Three numbers matches choice (C).

Alternative: Tool #4 (Introduce a Variable): write $N = 300k$ since $N$ must be a multiple of $300$. The condition $1000 < 300k < 2000$ divides into $\tfrac{10}{3} < k < \tfrac{20}{3}$, i.e., $3.33\ldots < k < 6.66\ldots$. The integer values of $k$ in that range are $4, 5, 6$ — three of them — confirming (C).

CCSS standards used (min grade 6)

6.NS.B.4 Find the greatest common factor and least common multiple of two whole numbers (Collapsing the three divisibility conditions into a single LCM condition and computing $\operatorname{lcm}(15, 20, 25) = 2^2 \cdot 3 \cdot 5^2 = 300$ from prime factorizations.)
4.OA.B.4 Find all factor pairs and recognize multiples in the range 1-100 (Listing multiples of $300$ in order ($900, 1200, 1500, 1800, 2100$) and identifying which lie strictly between $1000$ and $2000$.)
4.OA.A.3 Solve multi-step word problems using the four operations with whole numbers (Counting the qualifying multiples $\{1200, 1500, 1800\}$ to arrive at the final answer of $3$.)

⭐ Several divisibility rules at once is really one LCM rule in disguise. Find the LCM, then list its multiples inside the range and count.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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