AMC 8 · 2016 · #20
Grade 6 number-theoryProblem
The least common multiple of a and b is 12, and the least common multiple of b and c is 15. What is the least possible value of the least common multiple 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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