Prime Factorisation, HCF and LCM

Every number in the product must be a prime number, and 1 is not prime.

The two numbers on the branches below the box multiply to give the number in the box.

Keep dividing 72 by 2 until you can't, then divide by 3.

List the factors of each number and find the biggest one that is in both lists.

List the multiples of each number until you find one that is in both lists.

Every number in the tree is the product of the two numbers on the branches below it.

Part (a)

1. 15=3\times5, so the missing number is 5.
2. 5 is prime, so the branch stops there.

Part (b)

1. The primes at the ends of the branches are 2, 5, 3 and 5.
2. 150=2\times3\times5\times5=2\times3\times5^2, so n=2.

Divide 168 by 2 as many times as you can, then divide what is left by 3 and by 7.

Part (a)

1. 168\div2=84, 84\div2=42, 42\div2=21.
2. 21 is odd, so 2 appears 3 times and a=3.

Part (b)

1. 21=3\times7, so 3 appears once.
2. 3=3^1, so b=1.

Part (c)

1. 168=2\times2\times2\times3\times7=2^3\times3\times7.
2. 7 appears once, so c=1.

Use only the primes that are in both A and B, each with the smaller power.

1. 2 is in both: the smaller power is 2^1.
2. 3 is in both: the smaller power is 3^1.
3. 7 and 5 are each in only one of them, so leave them out.
4. HCF =2\times3=6.

Use every prime that appears in A or B, each with the bigger power.

1. The primes are 2, 3 and 5.
2. Bigger powers: 2^3, 3^1 and 5^1.
3. LCM =2^3\times3\times5=8\times15=120.

Write each number as a product of its prime factors first.

1. 132=2\times2\times3\times11=2^2\times3\times11.
2. 198=2\times3\times3\times11=2\times3^2\times11.
3. The prime factors in both are 2, 3 and 11.
4. HCF =2\times3\times11=66.

You need the lowest common multiple of 12 and 20.

1. Multiples of 12: 12, 24, 36, 48, 60, \ldots
2. Multiples of 20: 20, 40, 60, \ldots
3. The first number in both lists is 60.
4. They next flash together after 60 seconds.

The number of packs must divide exactly into both 48 and 72.

Part (a)

1. 48=2^4\times3 and 72=2^3\times3^2.
2. HCF =2^3\times3=24.
3. So the greatest number of packs is 24.

Part (b)

1. 72\div24=3.
2. There are 3 pencils in each pack.

Find the lowest common multiple of 16 and 20, then add that many minutes to 8:30 am.

1. Multiples of 16: 16, 32, 48, 64, 80, \ldots
2. Multiples of 20: 20, 40, 60, 80, \ldots
3. The LCM is 80 minutes, which is 1 hour 20 minutes.
4. 8:30 am + 1 hour 20 minutes = 9:50 am.

The side of a tile must divide exactly into both 96 and 72.

Part (a)

1. 96=2^5\times3 and 72=2^3\times3^2.
2. HCF =2^3\times3=24.
3. The largest tile has sides of 24 cm.

Part (b)

1. Along the length: 96\div24=4 tiles.
2. Along the width: 72\div24=3 tiles.
3. Number of tiles =4\times3=12.

A factor of a number can't be bigger than the number.

Part (a)

1. 72\div18=4 and 72\div24=3, so 72 is a common multiple, not a common factor.
2. 72 is the lowest common multiple (LCM) of 18 and 24.

Part (b)

1. Factors of 18: 1, 2, 3, 6, 9, 18.
2. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
3. The biggest factor in both lists is 6.

Write each number as a product of its prime factors, or list multiples of 108 until one is also a multiple of 90.

1. 90=2\times3^2\times5.
2. 108=2^2\times3^3.
3. Use each prime with its bigger power: 2^2, 3^3 and 5.
4. LCM =2^2\times3^3\times5=4\times27\times5=540.

A factor of N can only use the primes in N, with powers no bigger than the powers in N.

Part (a)

1. 2^3=8 and 5^2=25.
2. N=8\times3\times25=24\times25=600.

Part (b)

1. N has only three 2s, because of the 2^3.
2. 2^4\times3 needs four 2s, so it is not a factor of N.
3. Check: 2^4\times3=48 and 600\div48=12.5, which is not a whole number.

For the HCF use the smaller power of each prime; for the LCM use the bigger power.

Part (a)

1. Smaller powers: 2^2, 3^1 and 5^1.
2. HCF =2^2\times3\times5=4\times3\times5=60.

Part (b)

1. Bigger powers: 2^4, 3^3 and 5^2.
2. LCM =2^4\times3^3\times5^2=16\times27\times25.
3. 16\times27=432 and 432\times25=10\,800.

The length of a small piece must divide exactly into 36, 60 and 84.

Part (a)

1. 36=2^2\times3^2, 60=2^2\times3\times5 and 84=2^2\times3\times7.
2. The primes in all three are 2^2 and 3.
3. HCF =2^2\times3=12, so each piece is 12 cm.

Part (b)

1. 36\div12=3, 60\div12=5 and 84\div12=7.
2. Altogether 3+5+7=15 pieces.

The number of plates must be a multiple of 15 and the number of cups a multiple of 18.

Part (a)

1. Multiples of 18: 18, 36, 54, 72, 90, \ldots
2. 90 is the first of these that is also a multiple of 15, so the LCM is 90.
3. Packs of plates: 90\div15=6.

Part (b)

1. He buys 90 cups.
2. Packs of cups: 90\div18=5.