Enough beads to build both

The idea

In the prime factorisation method, the LCM of given numbers is the product of the greatest power of each prime factor involved in the numbers.

A number both can build

We want a number that both 6 = 2 × 3 and 20 = 2² × 5 divide into. It must contain every bead of 6 and every bead of 20.

Which is the smaller such number?

• 60
• 120
• 26

Take enough beads

For each prime, take enough beads to build every number. Then check your LCM.

Greatest power of every prime

The LCM is the product of the greatest power of each prime factor involved in the numbers.

So LCM(6, 20) = 2^2 × 3^1 × 5^1 = 60. Note the difference: the HCF takes the smallest power of the common primes; the LCM takes the greatest power of all the primes.

• Prime | In 6 | In 20 | Greatest power
• 2 | 2^1 | 2^2 | 2^2
• 3 | 3^1 | none | 3^1
• 5 | none | 5^1 | 5^1

Notes

The LCM is the product of the greatest power of each prime factor involved in the numbers. Example: LCM(6, 20) = 2^2 × 3 × 5 = 60.

Check yourself

Find the LCM of 12 = 2² × 3 and 18 = 2 × 3².

Answer: 36

Greatest powers: 2^2 and 3^2. LCM = 4 × 9 = 36.

8 = 2³ and 12 = 2² × 3. Which prime powers must the LCM contain?

15 = 3 × 5 and 28 = 2² × 7 share no prime. What is their LCM?

Answer: 420

No common primes, so we take them all: 2^2 × 3 × 5 × 7 = 420, which is 15 × 28.

Find the LCM of 6, 72 and 120, given 6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5.

Answer: 360

Greatest powers: 2^3, 3^2 and 5^1. LCM = 8 × 9 × 5 = 360.

• 2^2 and 3. The LCM must also build 8, which needs 2^3.
• 2^3 and 3 — correct. Yes! 8 needs 2^3 and 12 needs 3. LCM = 8 × 3 = 24.
• 2^3 only. Then 12 could not be built: it needs a 3.
• 2^5 and 3. We take the greatest power of 2, not the sum of the powers.