GCF and LCM: Additional Practice (Set 1)

Write $8 = 2^3$, $9 = 3^2$, and $12 = 2^2 \times 3$. Take the highest power of each prime across all three: $2^3$ and $3^2$.

$\operatorname{LCM}(8, 9, 12) = 2^3 \times 3^2 = 8 \times 9 = 72$

Each of $8$, $9$, and $12$ divides $72$, and nothing smaller works for all three.

The number of bouquets must divide $60$, $36$, and $24$, as large as possible, so this is the GCF of all three. Write $60 = 2^2 \times 3 \times 5$, $36 = 2^2 \times 3^2$, $24 = 2^3 \times 3$, then take the lowest power of each shared prime.

$\operatorname{GCF}(60, 36, 24) = 2^2 \times 3 = 12$

So $12$ bouquets, each with $5$ roses, $3$ lilies, and $2$ irises.

Write $20 = 2^2 \times 5$ and $50 = 2 \times 5^2$. Take the highest power of each prime: $2^2$ and $5^2$.

$\operatorname{LCM}(20, 50) = 2^2 \times 5^2 = 4 \times 25 = 100$

The value $10$ is the GCF, and $1000 = 20 \times 50$ is the plain product.

Rearrange the product rule for two numbers, $\operatorname{GCF}(a, b) \times \operatorname{LCM}(a, b) = a \times b$, to solve for the LCM.

$\operatorname{LCM}(a, b) = (a \times b) \div \operatorname{GCF}(a, b) = 360 \div 6 = 60$

So the LCM is $60$, regardless of which specific pair multiplies to $360$ with GCF $6$.

Factor each: $91 = 7 \times 13$ and $143 = 11 \times 13$. The only shared prime is $13$.

$\operatorname{GCF}(91, 143) = 13$

These numbers look like they might be prime, but each splits, and $13$ is the factor they have in common.

Write $21 = 3 \times 7$, $28 = 2^2 \times 7$, and $14 = 2 \times 7$. Take the highest power of each prime: $2^2$, $3$, $7$.

$\operatorname{LCM}(21, 28, 14) = 2^2 \times 3 \times 7 = 4 \times 3 \times 7 = 84$

The prime $7$ appears once in each, so one copy is enough; $2$ needs its highest power $2^2$ from $28$.

Take the lower power of each shared prime: $2^3$ (the lower of $2^5$ and $2^3$) and $3^2$ (lower of $3^2$ and $3^4$). The prime $7$ is in only one number, so it drops out.

$\operatorname{GCF} = 2^3 \times 3^2 = 8 \times 9 = 72$

Working from the prime powers avoids multiplying the large numbers out first.

Take the higher power of each prime: $2^5$ (the higher of $2^5$ and $2^3$), $3^4$ (higher of $3^2$ and $3^4$), and the $7$ that appears in the second number.

$\operatorname{LCM} = 2^5 \times 3^4 \times 7$

You choose the larger power for each prime and include every prime that shows up; you never add exponents.

Write $120 = 2^3 \times 3 \times 5$ and $200 = 2^3 \times 5^2$. Shared primes are $2$ and $5$; take the lower power of each: $2^3$ and $5^1$.

$\operatorname{GCF}(120, 200) = 2^3 \times 5 = 8 \times 5 = 40$

The prime $3$ is only in $120$, so it is not part of the common factor.

All three meet at a common multiple of $4$, $6$, and $8$, and next at the least one. Write $4 = 2^2$, $6 = 2 \times 3$, $8 = 2^3$, then take the highest power of each prime.

$\operatorname{LCM}(4, 6, 8) = 2^3 \times 3 = 8 \times 3 = 24$

So after $24$ minutes all three are at the start together. The value $2$ is the GCF, not the answer.

Write $45 = 3^2 \times 5$ and $75 = 3 \times 5^2$. Take the highest power of each prime: $3^2$ and $5^2$.

$\operatorname{LCM}(45, 75) = 3^2 \times 5^2 = 9 \times 25 = 225$

The value $15$ is the GCF, and $3375 = 45 \times 75$ is the plain product.

Write $168 = 2^3 \times 3 \times 7$ and $210 = 2 \times 3 \times 5 \times 7$. Shared primes are $2, 3, 7$; take the lower power of each: $2^1$, $3^1$, $7^1$.

$\operatorname{GCF}(168, 210) = 2 \times 3 \times 7 = 42$

The prime $5$ is only in $210$, so it is left out of the common factor.