AMC 8 · 2016 · #20
Easy mode Grade 6Problem
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?
Pick an answer.
AMC 8 2016 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: 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
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 $5$.
💡 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.
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 $b = 3$.
💡 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.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.
💡 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.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.
💡 When $b$ shares nothing, every prime gets pushed entirely into $a$ or $c$, blowing up $\text{lcm}(a, c)$.
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).
💡 Only two cases survived the $b$-filter, so the brute compare is just "pick the smaller of two numbers".
6.EE.A.1 Write $12$ and $15$ as products of primes. This tells us which primes can appear 6.NS.B.4 Narrow down $b$. Since $b$ is a factor of $\text{lcm}(a, b) = 12$ and also a fac 6.NS.B.4 Case $b = 3$. With $b$ already supplying the factor of $3$, $a$ only needs to ca 6.NS.B.4 Case $b = 1$. Now $b$ supplies no primes at all, so $a$ alone must make $\text{l 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.4Find 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.1Write 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.4Find 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 $\text{lcm}(4, 5) = 20$!
⭐ 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 $\text{lcm}(4, 5) = 20$!
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