GCF and LCM: Additional Practice (Set 2)

Write $72 = 2^3 \times 3^2$ and $120 = 2^3 \times 3 \times 5$. Shared primes are $2$ and $3$; take the lower power of each: $2^3$ and $3^1$.

$\operatorname{GCF}(72, 120) = 2^3 \times 3 = 8 \times 3 = 24$

The prime $5$ is only in $120$, so it drops out of the common factor.

Write $54 = 2 \times 3^3$ and $144 = 2^4 \times 3^2$. Shared primes are $2$ and $3$; take the lower power of each: $2^1$ and $3^2$.

$\operatorname{GCF}(54, 144) = 2 \times 3^2 = 2 \times 9 = 18$

The prime $3$ appears three times in $54$ and twice in $144$, so the GCF keeps the smaller count, two.

Write $72 = 2^3 \times 3^2$ and $120 = 2^3 \times 3 \times 5$. Take the highest power of each prime: $2^3$, $3^2$, and $5$.

$\operatorname{LCM}(72, 120) = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$

The value $8640 = 72 \times 120$ is the plain product, far larger than the least common multiple.

Write $54 = 2 \times 3^3$ and $144 = 2^4 \times 3^2$. Take the highest power of each prime: $2^4$ and $3^3$.

$\operatorname{LCM}(54, 144) = 2^4 \times 3^3 = 16 \times 27 = 432$

The value $7776 = 54 \times 144$ is the plain product, much larger than the least common multiple.

Write $15 = 3 \times 5$, $18 = 2 \times 3^2$, and $20 = 2^2 \times 5$. Take the highest power of each prime across all three: $2^2$, $3^2$, $5$.

$\operatorname{LCM}(15, 18, 20) = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$

The rule extends past two numbers: still take the highest power of every prime that appears.

By the product rule for two numbers, $a \times b = \operatorname{GCF}(a, b) \times \operatorname{LCM}(a, b) = 15 \times 90 = 1350$. Divide by $a = 45$.

$b = 1350 \div 45 = 30$

Check: $\operatorname{GCF}(45, 30) = 15$ and $\operatorname{LCM}(45, 30) = 90$, as required.

The row size must divide both $36$ and $48$, and we want it as large as possible, so this is the GCF. Write $36 = 2^2 \times 3^2$ and $48 = 2^4 \times 3$, then take the lower power of each shared prime.

$\operatorname{GCF}(36, 48) = 2^2 \times 3 = 12$

So each row holds $12$ flowers, giving $3$ rows of tulips and $4$ rows of daffodils.

Numbers with GCF $1$ share no prime. Factor $15 = 3 \times 5$ and $28 = 2^2 \times 7$: they share no prime at all.

$\operatorname{GCF}(15, 28) = 1$

The other pairs each share a factor: $14$ and $21$ share $7$, $18$ and $24$ share $6$, and $20$ and $30$ share $10$.

Write $48 = 2^4 \times 3$, $72 = 2^3 \times 3^2$, and $120 = 2^3 \times 3 \times 5$. Take the lowest power of each prime shared by all three: $2^3$ and $3^1$.

$\operatorname{GCF}(48, 72, 120) = 2^3 \times 3 = 8 \times 3 = 24$

The prime $5$ is missing from two of the numbers, so it cannot be in the common factor.

They flash together at a common multiple of $9$, $12$, and $18$, and next at the least one. Write $9 = 3^2$, $12 = 2^2 \times 3$, $18 = 2 \times 3^2$, then take the highest power of each prime.

$\operatorname{LCM}(9, 12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36$

So all three flash together again after $36$ seconds. The value $3$ is the GCF, not the answer.

Write $24 = 2^3 \times 3$ and $36 = 2^2 \times 3^2$. Take the highest power of each prime: $2^3$ and $3^2$.

$\operatorname{LCM}(24, 36) = 2^3 \times 3^2 = 8 \times 9 = 72$

The value $12$ is the GCF, and $864 = 24 \times 36$ is the plain product.

The GCF is always a factor of both numbers and is at most the smaller one. If it equals one of the numbers, that number must be the smaller, and it divides the larger evenly.

$\operatorname{GCF}(6, 18) = 6 \;\Rightarrow\; 6 \text{ divides } 18$

For example $\operatorname{GCF}(6, 18) = 6$ because $6$ divides $18$. In that case the larger number is also the LCM.