LCM and GCF Word Problems 01

24 days later, on Saturday, November 7, 2026.

The machines are inspected every 6, 8, and 12 days. We need the first day when all three schedules meet again.

$$ \begin{aligned} 6 &= 2 \times 3 \\ 8 &= 2^3 \\ 12 &= 2^2 \times 3 \end{aligned} $$

Use the needed factors \(2^3\) and \(3\):

$$ 2^3 \times 3 = 24 $$

So all three machines are inspected together again 24 days after October 14, 2026.

October 14, 2026 plus 24 days is Saturday, November 7, 2026.

36 days later, on a Tuesday.

The practices happen every 4, 6, and 9 days. We need the first day when all three schedules meet again.

$$ \begin{aligned} 4 &= 2^2 \\ 6 &= 2 \times 3 \\ 9 &= 3^2 \end{aligned} $$

Use the needed factors \(2^2\) and \(3^2\):

$$ 2^2 \times 3^2 = 36 $$

They will all practice again after 36 days.

Since \(35\) days is exactly 5 weeks, \(36\) days after Monday is Tuesday.

60 days.

The deliveries arrive every 10, 15, and 20 days. We need the first time all three arrive together again.

$$ \begin{aligned} 10 &= 2 \times 5 \\ 15 &= 3 \times 5 \\ 20 &= 2^2 \times 5 \end{aligned} $$

Use the needed factors \(2^2\), \(3\), and \(5\):

$$ 2^2 \times 3 \times 5 = 60 $$

All three deliveries arrive together again after 60 days.

24 packages. Each package has 2 notebooks, 3 pencils, and 4 erasers.

To make the greatest number of identical packages with nothing left over, find the biggest number that divides 48, 72, and 96.

$$ \begin{aligned} 48 &= 2^4 \times 3 \\ 72 &= 2^3 \times 3^2 \\ 96 &= 2^5 \times 3 \end{aligned} $$

The shared factors are \(2^3\) and \(3\):

$$ 2^3 \times 3 = 24 $$

So the teacher can make 24 packages.

Each package gets:

$$ \begin{aligned} 48 \div 24 &= 2 \text{ notebooks} \\ 72 \div 24 &= 3 \text{ pencils} \\ 96 \div 24 &= 4 \text{ erasers} \end{aligned} $$

60 cm, or 0.6 m, is the longest possible length for each piece.

First change all the ribbon lengths to centimeters.

$$ \begin{aligned} 1.2\text{ m} &= 120\text{ cm} \\ 1.8\text{ m} &= 180\text{ cm} \\ 2.4\text{ m} &= 240\text{ cm} \end{aligned} $$

We need the longest length that can divide all three ribbon lengths with no leftovers.

$$ \begin{aligned} 120 &= 2^3 \times 3 \times 5 \\ 180 &= 2^2 \times 3^2 \times 5 \\ 240 &= 2^4 \times 3 \times 5 \end{aligned} $$

The shared factors are \(2^2\), \(3\), and \(5\):

$$ 2^2 \times 3 \times 5 = 60 $$

The longest equal pieces are 60 cm, which is 0.6 m.

12 packages. Each package has 3 sardine cans, 5 instant-noodle packages, and 7 UHT milk cartons.

To make the greatest number of identical donation packages with nothing left over, find the biggest number that divides 36, 60, and 84.

$$ \begin{aligned} 36 &= 2^2 \times 3^2 \\ 60 &= 2^2 \times 3 \times 5 \\ 84 &= 2^2 \times 3 \times 7 \end{aligned} $$

The shared factors are \(2^2\) and \(3\):

$$ 2^2 \times 3 = 12 $$

So the group can make 12 packages.

Each package gets:

$$ \begin{aligned} 36 \div 12 &= 3 \text{ sardine cans} \\ 60 \div 12 &= 5 \text{ instant-noodle packages} \\ 84 \div 12 &= 7 \text{ UHT milk cartons} \end{aligned} $$

18 days.

The visits repeat every 6 days and 9 days. We need the first time both visits happen on the same day again.

$$ \begin{aligned} 6 &= 2 \times 3 \\ 9 &= 3^2 \end{aligned} $$

Use the needed factors \(2\) and \(3^2\):

$$ 2 \times 3^2 = 18 $$

Both classes will visit the library together again after 18 days.

8 bags. Each bag has 3 apples and 5 oranges.

To make the greatest number of identical bags with nothing left over, find the biggest number that divides 24 and 40.

$$ \begin{aligned} 24 &= 2^3 \times 3 \\ 40 &= 2^3 \times 5 \end{aligned} $$

The shared factor is \(2^3\):

$$ 2^3 = 8 $$

So there can be 8 bags.

Each bag gets:

$$ \begin{aligned} 24 \div 8 &= 3 \text{ apples} \\ 40 \div 8 &= 5 \text{ oranges} \end{aligned} $$

30 days.

The plants are watered every 10 days and 15 days. We need the first day when both watering schedules meet again.

$$ \begin{aligned} 10 &= 2 \times 5 \\ 15 &= 3 \times 5 \end{aligned} $$

Use the needed factors \(2\), \(3\), and \(5\):

$$ 2 \times 3 \times 5 = 30 $$

Both plants are watered on the same day again after 30 days.

18 friends. Each friend gets 2 star stickers and 3 heart stickers.

To share equally with as many friends as possible, find the biggest number that divides 36 and 54.

$$ \begin{aligned} 36 &= 2^2 \times 3^2 \\ 54 &= 2 \times 3^3 \end{aligned} $$

The shared factors are \(2\) and \(3^2\):

$$ 2 \times 3^2 = 18 $$

Wait: \(18\) divides 36 and 54, so the greatest number of friends is 18.

Each friend gets:

$$ \begin{aligned} 36 \div 18 &= 2 \text{ star stickers} \\ 54 \div 18 &= 3 \text{ heart stickers} \end{aligned} $$

24 seconds.

The signs flash every 8 seconds and 12 seconds. We need the first time both signs flash together again.

$$ \begin{aligned} 8 &= 2^3 \\ 12 &= 2^2 \times 3 \end{aligned} $$

Use the needed factors \(2^3\) and \(3\):

$$ 2^3 \times 3 = 24 $$

Both signs flash together again after 24 seconds.

84 days later, on Sunday.

The restocks repeat every 12, 21, and 28 days.

$$ \begin{aligned} 12 &= 2^2 \times 3 \\ 21 &= 3 \times 7 \\ 28 &= 2^2 \times 7 \end{aligned} $$

Use the needed factors \(2^2\), \(3\), and \(7\):

$$ 2^2 \times 3 \times 7 = 84 $$

All three items are restocked together again after 84 days.

Since \(84\) days is exactly 12 weeks, the weekday stays the same. If today is Sunday, the next shared restock is also on Sunday.

18 boxes. Each box has 4 crayons, 5 markers, and 6 colored pencils.

To make the greatest number of identical boxes, find the biggest number that divides 72, 90, and 108.

$$ \begin{aligned} 72 &= 2^3 \times 3^2 \\ 90 &= 2 \times 3^2 \times 5 \\ 108 &= 2^2 \times 3^3 \end{aligned} $$

The shared factors are \(2\) and \(3^2\):

$$ 2 \times 3^2 = 18 $$

So there can be 18 boxes.

Each box gets:

$$ \begin{aligned} 72 \div 18 &= 4 \text{ crayons} \\ 90 \div 18 &= 5 \text{ markers} \\ 108 \div 18 &= 6 \text{ colored pencils} \end{aligned} $$

180 minutes, or 3 hours.

The lane rotations repeat every 20, 30, and 45 minutes.

$$ \begin{aligned} 20 &= 2^2 \times 5 \\ 30 &= 2 \times 3 \times 5 \\ 45 &= 3^2 \times 5 \end{aligned} $$

Use the needed factors \(2^2\), \(3^2\), and \(5\):

$$ 2^2 \times 3^2 \times 5 = 180 $$

The lane rotations line up again after 180 minutes, which is 3 hours.

1.5 meters is the longest possible length.

Convert the rope lengths to centimeters.

$$ \begin{aligned} 4.5\text{ m} &= 450\text{ cm} \\ 6\text{ m} &= 600\text{ cm} \\ 7.5\text{ m} &= 750\text{ cm} \end{aligned} $$

Find the greatest length that divides 450, 600, and 750.

$$ \begin{aligned} 450 &= 2 \times 3^2 \times 5^2 \\ 600 &= 2^3 \times 3 \times 5^2 \\ 750 &= 2 \times 3 \times 5^3 \end{aligned} $$

The shared factors are \(2\), \(3\), and \(5^2\):

$$ 2 \times 3 \times 5^2 = 150 $$

The longest equal pieces are 150 cm, or 1.5 m.

120 minutes, or 2 hours. If they leave together at 6:00 a.m., they leave together again at 8:00 a.m.

The buses leave every 24, 40, and 60 minutes.

$$ \begin{aligned} 24 &= 2^3 \times 3 \\ 40 &= 2^3 \times 5 \\ 60 &= 2^2 \times 3 \times 5 \end{aligned} $$

Use the needed factors \(2^3\), \(3\), and \(5\):

$$ 2^3 \times 3 \times 5 = 120 $$

The buses actually leave together again after 120 minutes, or 2 hours.

Starting from 6:00 a.m., that is 8:00 a.m.

36 packs. Each pack has 4 stickers, 5 bookmarks, and 7 postcards.

Find the greatest number that divides 144, 180, and 252.

$$ \begin{aligned} 144 &= 2^4 \times 3^2 \\ 180 &= 2^2 \times 3^2 \times 5 \\ 252 &= 2^2 \times 3^2 \times 7 \end{aligned} $$

The shared factors are \(2^2\) and \(3^2\):

$$ 2^2 \times 3^2 = 36 $$

So the students can make 36 packs.

Each pack gets:

$$ \begin{aligned} 144 \div 36 &= 4 \text{ stickers} \\ 180 \div 36 &= 5 \text{ bookmarks} \\ 252 \div 36 &= 7 \text{ postcards} \end{aligned} $$

420 minutes, or 7 hours. Starting at 7:30 a.m., they happen together again at 2:30 p.m.

The events repeat every 28, 35, and 60 minutes.

$$ \begin{aligned} 28 &= 2^2 \times 7 \\ 35 &= 5 \times 7 \\ 60 &= 2^2 \times 3 \times 5 \end{aligned} $$

Use the needed factors \(2^2\), \(3\), \(5\), and \(7\):

$$ 2^2 \times 3 \times 5 \times 7 = 420 $$

So all three happen together again after 420 minutes, which is 7 hours.

Starting at 7:30 a.m., 7 hours later is 2:30 p.m.

48 kits. Each kit has 3 m of string, 4 m of ribbon, and 5 m of yarn.

Because each kit must contain a whole number of meters of each material, use the meter amounts: 144, 192, and 240.

$$ \gcd(144,192,240)=48 $$

So the greatest number of kits with whole-meter amounts in each kit is 48 kits.

Each kit gets:

$$ \begin{aligned} 144 \div 48 &= 3\text{ m of string} \\ 192 \div 48 &= 4\text{ m of ribbon} \\ 240 \div 48 &= 5\text{ m of yarn} \end{aligned} $$

The teams meet again after 180 minutes. They can make 12 crates, with 8 gloves, 15 masks, and 21 trash bags in each crate.

This problem has two parts.

First, find when the teams meet again. Their schedules repeat every 30, 45, and 60 minutes.

$$ \begin{aligned} 30 &= 2 \times 3 \times 5 \\ 45 &= 3^2 \times 5 \\ 60 &= 2^2 \times 3 \times 5 \end{aligned} $$

Use the needed factors \(2^2\), \(3^2\), and \(5\):

$$ 2^2 \times 3^2 \times 5 = 180 $$

So all three teams meet again after 180 minutes.

Next, find the greatest number of identical supply crates from 96 gloves, 180 masks, and 252 trash bags.

$$ \begin{aligned} 96 &= 2^5 \times 3 \\ 180 &= 2^2 \times 3^2 \times 5 \\ 252 &= 2^2 \times 3^2 \times 7 \end{aligned} $$

The shared factors are \(2^2\) and \(3\):

$$ 2^2 \times 3 = 12 $$

They can make 12 crates.

Each crate gets:

$$ \begin{aligned} 96 \div 12 &= 8 \text{ gloves} \\ 180 \div 12 &= 15 \text{ masks} \\ 252 \div 12 &= 21 \text{ trash bags} \end{aligned} $$