AMC 8 · 2016 · #20

Easy mode Grade 6

Problem

The least common multiple (lcm) of two numbers is the smallest number that both of them divide evenly into.

We have three whole numbers $a$ , $b$ , and $c$ . We are told two facts:

The lcm of $a$ and $b$ is $12$ .
The lcm of $b$ and $c$ is $15$ .

Among all choices of $a$ , $b$ , $c$ that fit these two facts, what is the smallest possible value of the lcm of $a$ and $c$ ?

$\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180$

Pick an answer.

(A)

(B)

(C)

(D)

120

(E)

180

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Toolkit + CCSS Solution

Understand

Restated: Three positive integers $a$, $b$, $c$ satisfy $\text{lcm}(a, b) = 12$ and $\text{lcm}(b, c) = 15$. Find the smallest possible value of $\text{lcm}(a, c)$.

Givens: $\text{lcm}(a, b) = 12 = 2^2 \times 3$; $\text{lcm}(b, c) = 15 = 3 \times 5$; $a$, $b$, $c$ are positive integers; Answer choices: (A) $20$, (B) $30$, (C) $60$, (D) $120$, (E) $180$

Unknowns: The minimum possible value of $\text{lcm}(a, c)$

Understand

Restated: Three positive integers $a$, $b$, $c$ satisfy $\text{lcm}(a, b) = 12$ and $\text{lcm}(b, c) = 15$. Find the smallest possible value of $\text{lcm}(a, c)$.

Givens: $\text{lcm}(a, b) = 12 = 2^2 \times 3$; $\text{lcm}(b, c) = 15 = 3 \times 5$; $a$, $b$, $c$ are positive integers; Answer choices: (A) $20$, (B) $30$, (C) $60$, (D) $120$, (E) $180$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #6 Guess and Check

Tool #7 (Identify Subproblems) is the natural fit. The two LCM conditions share the variable $b$, so we split the work in three pieces — first pin down what $b$ can be (it must divide both $12$ and $15$), then for each choice of $b$ find the smallest legal $a$ and the smallest legal $c$ separately, then take $\text{lcm}(a, c)$. Tool #6 (Guess and Check) handles the tiny case-list — only $b = 1$ and $b = 3$ survive — so we just test both and keep the winner. We avoid Tool #13 (Algebra) here because the search space is two cases, not an equation.

Execute — Answer: A

#7 Identify Subproblems 6.EE.A.1 Step 1

Write $12$ and $15$ as products of primes.
This tells us which primes can appear in $a$, $b$, $c$ at all: only $2$, $3$, and $5$.

$$12 = 2^2 \times 3, \;\; 15 = 3 \times 5$$

💡 Splitting a number into its prime building blocks is the first move whenever the question is about $\text{lcm}$ or $\gcd$ — each prime can be handled on its own.

#7 Identify Subproblems 6.NS.B.4 Step 2

Narrow down $b$.
Since $b$ is a factor of $\text{lcm}(a, b) = 12$ and also a factor of $\text{lcm}(b, c) = 15$, it must divide both.
The largest number that divides both $12$ and $15$ is $\gcd(12, 15) = 3$, so $b$ is a divisor of $3$.
That leaves only $b = 1$ or $b = 3$.

$$b \mid 12 \text{ and } b \mid 15 \Rightarrow b \mid \gcd(12, 15) = 3 \Rightarrow b \in \{1, 3\}$$

💡 If $b$ tried to use the prime $2$ it would force $\text{lcm}(b, c)$ to be even, but $15$ is odd. The same logic rules $5$ out of $b$ via $12$.

#6 Guess and Check 6.NS.B.4 Step 3

Case $b = 3$.
With $b$ already supplying the factor of $3$, $a$ only needs to carry the missing $2^2$ for $\text{lcm}(a, b) = 12$, so the smallest $a$ is $4$.
Likewise $c$ only needs to carry the missing $5$ for $\text{lcm}(b, c) = 15$, so the smallest $c$ is $5$.
Check: $\text{lcm}(4, 3) = 12$ and $\text{lcm}(3, 5) = 15$ — both conditions hold.

$$b = 3:\;\; a = 4,\; c = 5 \;\Rightarrow\; \text{lcm}(a, c) = \text{lcm}(4, 5) = 20$$

💡 Loading $b$ with every shared prime ($3$ in this case) lets $a$ and $c$ stay as small as possible, and $\text{lcm}(a, c)$ stays small with them.

#6 Guess and Check 6.NS.B.4 Step 4

Case $b = 1$.
Now $b$ supplies no primes at all, so $a$ alone must make $\text{lcm}(a, b) = 12$, forcing $a = 12$, and $c$ alone must make $\text{lcm}(b, c) = 15$, forcing $c = 15$.
Then $\text{lcm}(12, 15) = 60$ — much larger.

$$b = 1:\;\; a = 12,\; c = 15 \;\Rightarrow\; \text{lcm}(a, c) = \text{lcm}(12, 15) = 60$$

💡 When $b$ shares nothing, every prime gets pushed entirely into $a$ or $c$, blowing up $\text{lcm}(a, c)$.

#7 Identify Subproblems 4.OA.B.4 Step 5

Compare the two cases and pick the smaller.
$20 < 60$, so the minimum is $20$, which matches choice (A).

$$\min(20, 60) = 20 \;\Rightarrow\; \textbf{(A)}$$

💡 Only two cases survived the $b$-filter, so the brute compare is just "pick the smaller of two numbers".

[1] #7 6.EE.A.1 Write $12$ and $15$ as products of primes. This tells us which primes can appear

[2] #7 6.NS.B.4 Narrow down $b$. Since $b$ is a factor of $\text{lcm}(a, b) = 12$ and also a fac

[3] #6 6.NS.B.4 Case $b = 3$. With $b$ already supplying the factor of $3$, $a$ only needs to ca

[4] #6 6.NS.B.4 Case $b = 1$. Now $b$ supplies no primes at all, so $a$ alone must make $\text{l

[5] #7 4.OA.B.4 Compare the two cases and pick the smaller. $20 < 60$, so the minimum is $20$, w

Review

Reasonableness: Sanity-check the answer prime by prime. $\text{lcm}(a, c)$ must be a multiple of every prime that $a$ or $c$ is forced to carry. Looking at $12 = 2^2 \times 3$, the factor $2^2$ has no home in $b$ (since $b \mid 15$ rules $2$ out of $b$), so $a$ must carry it — forcing $4 \mid \text{lcm}(a, c)$. Similarly, $5$ has no home in $b$, so $c$ must carry it — forcing $5 \mid \text{lcm}(a, c)$. Hence $\text{lcm}(a, c)$ is at least $\text{lcm}(4, 5) = 20$. Our construction $(a, b, c) = (4, 3, 5)$ achieves exactly $20$, so $20$ is both attainable and a hard lower bound. That matches (A).

Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: any valid $\text{lcm}(a, c)$ must be divisible by $4$ (the $2^2$ has nowhere else to go) and by $5$ (same argument). So it must be a multiple of $20$. Among the choices $20, 30, 60, 120, 180$, the values $30$ is not a multiple of $4$ — eliminated. The remaining smallest is $20$, which we already achieved with $(a, b, c) = (4, 3, 5)$, so the answer is (A).

CCSS standards used (min grade 6)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing the divisors of $3$ ($1$ and $3$) to enumerate the legal values of $b$, and comparing the two resulting LCMs ($20$ vs. $60$) to pick the smaller.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Writing $12 = 2^2 \times 3$ and $15 = 3 \times 5$ in exponent form so each prime can be tracked independently.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Using $\gcd(12, 15) = 3$ to pin down the legal values of $b$, then computing $\text{lcm}(4, 3) = 12$, $\text{lcm}(3, 5) = 15$, and $\text{lcm}(4, 5) = 20$ to verify each case.)

⭐ This AMC 8 problem only needs the Grade 6 GCF/LCM idea — break $12$ and $15$ into primes, push the shared prime into $b$, and the leftover $4$ and $5$ give the smallest answer $\text{lcm}(4, 5) = 20$!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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