If you have to share or cut things into equal parts that are as large as possible, it is a GCD problem. If something repeats and you want to know when it lines up again, it is an LCM problem. For example, a 360 × 270 cm floor can be covered with square tiles 90 cm on a side, and two buses that leave every 12 and every 18 minutes are together again after 36 minutes.
| What the problem asks | Operation | Example |
|---|---|---|
| What is the largest size that fits a whole number of times into every quantity? | GCD | Tile for a 360 × 270 cm floor, 90 cm on a side |
| When do several cycles line up again? | LCM | Buses every 12 and 18 minutes, at 36 |
| What is the least common denominator? | LCM | 5/12 + 7/18 uses the denominator 36 |
| What can I divide by to simplify? | GCD | 18/24 divided by 6 gives 3/4 |
The words in a problem help, but they often mislead, so look at the logic. The GCD divides the numbers, so it is never larger than the smallest one. The LCM is a multiple of the numbers, so it is never smaller than the largest one. If your result breaks one of these two rules, you used the wrong operation.
Two solved problems
The first is a GCD problem and the second an LCM problem. In both you follow the same path: decide, calculate, answer with units, and check.
Worked example: tiles for a 360 × 270 cm floor
You want to cover a rectangular floor 360 cm by 270 cm with identical square tiles, as large as possible and without cutting any. How long is a side, and how many tiles do you need?
- Decide. The side must fit a whole number of times into both 360 and 270, and it must be as large as possible. That is the GCD of 360 and 270.
- Apply Euclid’s algorithm. 360 = 1 × 270 + 90 and 270 = 3 × 90 + 0, so the GCD is 90.
- Count the tiles. Along the length 360 ÷ 90 = 4 fit, and across the width 270 ÷ 90 = 3, so 4 × 3 = 12.
Result: tiles 90 cm on a side, 12 in all.
Check: 12 tiles of 90 × 90 = 8,100 cm² cover 97,200 cm², and 360 × 270 also gives 97,200. With 4 and 3 tiles per side there is no common divisor left, so no larger tile exists.
Worked example: two buses that meet again
Two routes leave the same stop, one every 12 minutes and the other every 18. They have just left together. When are they together again?
- Decide. You want the first moment that is a multiple of both 12 and 18. That is the LCM.
- Factor. 12 = 2² × 3 and 18 = 2 × 3². Take each prime with its highest exponent.
- Multiply. 2² × 3² = 36.
Result: after 36 minutes.
Check: 36 ÷ 12 = 3 and 36 ÷ 18 = 2, so both divisions are exact. No multiple of 18 smaller than 36 works, because 18 ÷ 12 is not a whole number. Multiplying 12 × 18 = 216 would give a time when they coincide, but not the first one.
If you need to review how each value is found, the guide on GCD vs. LCM compares them at a relaxed pace, and the one on the Euclidean algorithm explains why step 2 of the first example works.
With fractions, the same logic
Both ideas appear when you work with fractions. To add 5/12 + 7/18 you need the least common denominator, which is LCM(12, 18) = 36. Then 5/12 = 15/36, 7/18 = 14/36, and the sum is 29/36, which is already in lowest terms because 29 is prime.
To reduce a fraction, you divide the top and bottom by the GCD. The least common denominator calculator prepares fractions for adding, and the simplify fractions calculator applies the GCD. If a problem mixes sharing and fractions, also see the fraction word problems.
Five practice problems
They are mixed on purpose, without naming the operation. First decide whether the size fits inside the numbers or contains them, then calculate and write the answer with its unit.
Practice
- A rope of 45 m and another of 75 m are cut into equal pieces, as long as possible. How long is each piece, and how many pieces are there?
- One lighthouse flashes every 8 seconds and another every 12. They have just flashed together. When do they flash together again?
- There are 28 boys and 42 girls. Equal teams are formed, each with boys and girls, with nobody left out. What is the greatest number of teams, and who is on each?
- Three chores repeat every 4, 6 and 9 days. All three fall on today. In how many days do they fall together again?
- One clock chimes every 7 minutes and another every 9. When do they chime together again?
Show the answers
- It is a GCD problem. GCD(45, 75) = 15, so the pieces are 15 m long and you get 3 + 5 = 8.
- It is an LCM problem. LCM(8, 12) = 24, so they flash together after 24 seconds.
- It is a GCD problem. GCD(28, 42) = 14, so there are 14 teams, each with 2 boys and 3 girls.
- It is an LCM problem. LCM(4, 6, 9) = 36, so they fall together in 36 days.
- It is an LCM problem. 7 and 9 have no common factors, so LCM(7, 9) = 63 and they chime together after 63 minutes.
How to check your answer
Divide each number by the GCD you found. The divisions must be exact, and the quotients must have no common divisor. For an LCM, divide it by each number and check that the result is a whole number. Then make sure the answer says what was asked, because sometimes the GCD is the number of batches and other times it is the contents of each batch.
The greatest common divisor and least common multiple calculators accept several numbers and are useful for verifying your work. The prime factorization calculator gives the factors of each number, and the divisibility rules help you find them in your head.