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GCSE :: Prime Factorisation & LCM/HCF
@DrFrostMaths Objectives: Be able to find the prime factorisation of a number, and find the Lower Common Multiple or Highest Common Factor of two numbers. This resource is intended for both Foundation & Higher Tiers. Last modified: 21st January 2020
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STARTER :: Factorisation
Recall that a factor of a number is a number that it can be divided by without a remainder. For example, 3 is a factor of 12 as 12Γ·3 leaves no remainder. List the factors of the following numbers: 10: 1, 2, 5, 10 12: 1, 2, 3, 4, 5, 6, 12 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 ? ? ?
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10=πΓπ 42=πΓπΓπ 12=πΓπΓπ Prime Factorisation ? ? ? ? ? ? ? ?
Key Term: Product just means the numbers have been multiplied, e.g. the βproduct of 2 and 3β is 2Γ3=6 For reasons weβll see later, itβs often helpful to be able to write a number as a product of its prime factors. We only want to use prime numbers here. We can use them more than once if needed! 10=πΓπ 42=πΓπΓπ 12=πΓπΓπ ? ? Fro Tip: We usually give the prime factors in ascending order. ? ? ? ? ? ? To be more concise, we could write 2Γ2 as 2 2 , therefore 12= 2 2 Γ3
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360 36 10 9 4 2 5 3 3 2 2 Prime Factor Trees =2Γ2Γ2Γ3Γ3Γ5
This is not so bad when the numbers are small, but it would be harder to write larger numbers as a product of prime factors in our head! We can make this easier using a prime factor tree. Think of two numbers which multiple to give 360. Read all the numbers from the leaves. 360 =2Γ2Γ2Γ3Γ3Γ5 = 2 3 Γ 3 2 Γ5 36 10 We βbranchβ into these two numbers. Iβve used 360=36Γ10, but thereβs lots of pairs you could have chosen. 9 4 2 5 We can only βstopβ at a number of it is prime. But neither 36 nor 10 is prime, so we have to keep βbranching outβ of each. Fro Note: Trees in maths are upside down. 3 3 2 2 2 and 5 are both prime, so we circle them. These are βleavesβ of the tree and no longer branch out.
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Further Example At the leaves we have one 2, two 3s and three 5s. 2250 =2Γ 3 2 Γ 5 3 225 10 45 5 2 5 9 5 3 3
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1350=2Γ 3 3 Γ 5 2 Test Your Understanding ? Some Possible Trees ?
Using a tree, find the prime factorisation of 1350. When done, try coming up with more trees. What do you notice about the final result in each case? 1350=2Γ 3 3 Γ 5 2 ? Some Possible Trees ? 1350 1350 5 270 10 135 90 3 2 5 5 27 3 3 30 9 5 6 3 3 2 3 We always end up with the same leaves each time, and hence the same factorisation. Fundamental Law of Arithmetic/Unique Factorisation Theorem: Every positive integer can be uniquely expressed as a product of primes.
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EXTENSION :: Dealing with numbers in power form
Sometimes you might have a number with powers, but the base (the big number) is not prime. How would you prime factorise this? What if a base was repeated? 10 3 = 2Γ =2Γ5Γ2Γ5Γ2Γ = 2 3 Γ 5 3 ? 2 3 Γ 2 4 = 2Γ2Γ2 Γ 2Γ2Γ2Γ2 = 2 7 ? ? ? ? Fro Note: This is an example of a βlaw of indicesβ. Quickfire Questions: N: 9 10 = π ππ 8 10 = π ππ = π πππ Γ π πππ 12 12 = π ππ Γ π ππ 1 8 5 = π π Γπ ππ 12Γ =πΓ π πππ Working: = 3 10 Γ 3 10 = 3 20 ? 6 5 = π π Γ π π = π πππ Γ π πππ = π ππ Γ π ππ 70 4 = π π Γ π π Γ π π 5 5 Γ 5 2 = π π 7 20 Γ 7 20 = π ππ ? ? ? ? ? ? ? ? ? ? ?
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Exercise 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? (on printed worksheet) 1 3
Write the following as a product of prime numbers (you will not need to use a tree). 35=πΓπ 8=πΓπΓπ= π π 30=πΓπΓπ 3 [IMC 2001 Q6] 2001=3Γ23Γ29. Which of the following numbers is also the product of exactly three distinct prime numbers? [ Β ]Β [ Β ]Β [ Β ]Β 91 [ Β ]Β 105 [ Β ]Β 330 Answer: 105 [Edexcel IGCSE(9-1) Jan 2019(R) 1F Q24b, Jan 2019(R) 1H Q9b] π=480Γ 10 9 Write π as a product of powers of its prime factors. π΅= π ππ ΓπΓ π ππ ? ? ? ? By drawing a tree of otherwise, find prime factorisations (in index form) for the following numbers. 56= π π Γπ 75=πΓ π π 126=πΓ π π Γπ 204= π π ΓπΓππ 168= π π ΓπΓπ 180= π π Γ π π Γπ 300= π π ΓπΓ π π 825=πΓ π π Γππ 792= π π Γ π π Γππ 2 4 ? ? ? ? ? ? ? ? ? ?
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Exercise 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? (on printed worksheet) 5 7 N1 6
[Edexcel IGCSE May2014(R)-4H Q8b Edited] Given that 240= 2 4 Γ3Γ5 and that 252= 2 2 Γ 3 2 Γ7 and that π¦=240Γ252 write π¦ as a product of powers of its prime factors. π= π π Γ π π ΓπΓπ 7 Prime factorise the following: 10 000=π π π = π π Γ π π = π ππ = π ππ Γ π ππ Γ 8 6 = π ππ = π πππ Γ π πππ Suppose 1 was considered to be a prime number. Explain why this violates the Fundamental Law of Arithmetic. For example, 6 could be expressed as πΓπ or πΓπΓπ or πΓπΓπΓπ. But FLA states there is a unique factorisation for each integer. Thus 1 is not prime. [TMC Regional 2012 Q4] Find the sum of all numbers less than 120 which are the product of exactly three different prime factors. Solution: 717 5 ? ? ? ? ? N1 ? ? 6 Put in prime factorised form: 3 5 Γ 3 6 = π ππ Γ 3 3 Γ 2 4 Γ 3 5 = π π Γ π π = π π Γ π π = π π Γπ π π = π ππ Γ ππ ππ 6Γ 3 20 =πΓ π ππ ? ? ? ? N2 ? ? ?
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Starter List out the factors of 8 and 12.
What factors do they have in common? Factors of 8: 1, 2, 4, 8 Factors of 12: 1, 2, 3, 4, 6, 12 Factors in common: 1, 2, 4 ? ? ? List out the some multiples of 8 and 12. What multiples do they have in common? Multiples of 8: 8, 16, 24, 32, 40, 48, β¦ Multiples of 12: 12, 24, 36, 48, 60, 72, β¦ Multiples in common: 24, 48, 72, β¦ ? ? ?
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Lowest Common Multiple/Highest Common Factor
Multiples of 8: 8, 16, 24, 32, β¦ Multiples of 12: 12, 24, 36, β¦ Lowest Common Multiple of 8 and 12: 24 Factors of 8: 1, 2, 4, 8 Factors of 12: 1, 2, 3, 4, 6, 12 Highest Common Factor of 8 and 12: 4 ? For small numbers, we can list out multiples of the larger number until we see a multiple of the smaller number. ? For small numbers, we can list out factors of each number and choose the greatest number which is common.
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Check Your Understanding
Fro Shortcut: Any multiple of 60 ends with a 0. Therefore the multiple of 72, in order to end with a 0, must be x5, x10, β¦ ? 1 πΏπΆπ 60,72 =πππ π»πΆπΉ 60,72 =ππ ? Fro Pro Shortcut: Any number which goes into 60 and 72 must also go into their difference! (i.e. 12) (This principle is the foundation of something called Euclidβs algorithm) ? 2 πΏπΆπ 12,21 =ππ π»πΆπΉ 12,21 =π ? I have to take two different pills on a regular basis to combat Frostitus. Pill A I take every 6 days. Pill B I take every 8 days. If I take both pills on January 1st, when is the next date I take both pills? The number of days that have passed have to be a multiple of 6 and of 8. π³πͺπ΄ π,π =ππ January 25th 3 ?
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But what about bigger numbers?
792, 378 Sometimes itβs not practical to use this method. Can we use the prime factorisation somehow? ? Prime Factorisation 792= 2 3 Γ 3 2 Γ11 378=2Γ 3 3 Γ7 ? Prime Factorisation
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But what about bigger numbers?
792= 2 3 Γ 3 2 Γ11 378=2Γ 3 3 Γ7 Exam Note: Both Foundation and Higher Tier students are expected to be able to do this. The βwhat wins, what losesβ method 792= 2 3 Γ Γ11 378=2 Γ 3 3 Γ7 Step 1: Align numbers so that each prime factor has its own column. π»πΆπΉ=2 Γ 3 2 ? In terms of the 2s, what do both and have in common as a factor? ? Step 2: To calculate HCF, see βwhat losesβ for each of the prime factors (i.e. lowest power, where βnothingβ always loses against βsomethingβ). ? πΏπΆπ= 2 3 Γ 3 3 Γ7Γ =16632 ? ? This time, what do both and 2 go into? ? Step 2: To calculate LCM, see βwhat winsβ.
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Quickfire Questions Find the LCM and HCF of each. The prime factors have already been lined up for you. 24 = 2 3 Γ =2 Γ 3 2 π»πΆπΉ=2 Γ3 πΏπΆπ= 2 3 Γ 3 2 Recap: βWhat losesβ ? ? 45 = Γ = 2 2 Γ3 π»πΆπΉ= πΏπΆπ= 2 2 Γ 3 2 Γ5 Recap: βWhat winsβ ? ? ? ? Fro Note: βSomethingβ always beats βnothingβ. 30 =2 Γ3Γ = Γ3 π»πΆπΉ=2 Γ3 πΏπΆπ= 2 2 Γ3Γ5 ? Fro Note: A βdrawβ counts as both a win and a lose. ?
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More Examples 2016= 2 5 Γ 3 2 Γ7 72 = 2 3 Γ 3 2 588= 2 2 Γ3Γ = 2 4 Γ7Γ11 ? ? π³πͺπ΄ ππππ,ππ = π π Γ π π Γπ=ππππ π―πͺπ ππππ,ππ = π π Γπ=ππ (note that if thereβs a βdrawβ, both win and both lose) Line numbers up: πππ = π π ΓπΓ π π ππππ= π π ΓπΓππ π³πͺπ΄ πππ,ππππ = π π ΓπΓ π π Γππ=πππππ π―πͺπ πππ,ππππ = π π Γπ=ππ
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Test Your Understanding
1936= π π Γ ππ π 792= π π Γ π π Γππ πΏπΆπ= π π Γ π π Γ ππ π =πππππ π»πΆπΉ= π π Γππ=ππ ? ? If you finishβ¦ ? 3675=πΓ π π Γ π π 875= π π Γπ ? ? πΏπΆπ=πΓ π π Γ π π =πππππ π»πΆπΉ= π π Γπ=πππ ?
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Exercise 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (on printed worksheet)
The K4 bus comes every 9 minutes. The K3 bus comes every 12 minutes. If they both come at 9am, at what time will they next arrive at the same time? 9:36am Year 7 has 48 pupils in it. Year 8 has 90 pupils in it. Each year group has to be divided into football teams, so that all teams (across both Year 7 and 8) are the same size, and no people are left over. Whatβs the largest team size? HCF(48,90) = 6 [Edexcel GCSE Nov2013-1F Q27i, Nov2013-1H Q7i] Rita is going to make some cheeseburgers for a party. She buys some packets of cheese slices and some boxes of burgers. There are 20 cheese slices in each packet. There are 12 burgers in each box. Rita buys exactly the same number of cheese slices and burgers. (a) How many packets of cheese slices and how many boxes of burgers does she buy? 3 and 5 (b) Rita wants to put one cheese slice and one burger into each bread roll. She wants to use all the cheese slices and all the burgers. How many bread rolls does Rita need? 60 4 Find the LCM and HCF of the following pairs of numbers (using any suitable method). 6 and 8 HCF = 2, LCM = 24 13 and 5 HCF = 1, LCM = 65 12 and 15 HCF = 3, LCM = 60 21 and 35 HCF = 7, LCM = 105 Find the LCM and HCF of the following numbers in prime factorised form. 2 2 Γ Γ ο LCM = π π Γ π π =πππ HCF = π π Γ π π =ππ 2Γ 3 2 Γ Γ ο LCM =πΓ π π Γπ=πππ HCF =πΓ π π =ππ 2Γ 3 2 Γ Γ5Γ7 ο LCM =πΓ π π ΓπΓπ=πππ HCF =πΓπ=ππ 3 3 Γ Γ ο LCM = π π Γ π π Γπ=ππππ HCF =π Find the LCM and HCF of the following pairs, by prime factorising the numbers first (you are welcome to use the βFACTβ button on your calculator). a) 36 and 378 LCM = 756, HCF = 18 b) 315 and 3675 LCM = 11025, HCF = 105 c) 72 and 66 LCM = 792, HCF = 6 d) 2880 and 792 LCM = 31680, HCF = 72 e) 375 and 325 LCM = 4875, HCF = 25 f) 252 and 2079 LCM = 8316, HCF = 63 1 ? ? ? 5 ? ? 2 ? ? 6 ? ? ? 3 ? ? ? ? ? ? ? ?
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Exercise 2 ? ? ? ? ? ? (on printed worksheet) 7 10 8 11 9 12
[JMC 2009 Q18] Six friends are having dinner together in their local restaurant. The first eats there every day, the second eats there every other day, the third eats there every third day, the fourth eats there every fourth day, the fifth every fifth day and the sixth eats there every sixth day. They agree to have a party the next time they all eat together there. In how many daysβ time is the party? 60 days [Edexcel IGCSE May2016-4H Q10] The highest common factor (HCF) of 140 and π₯ is 20 The lowest common multiple (LCM) of 140 and π₯ is Find the value of π₯. 60 [IMC 2013 Q15] I have a bag of coins. In it, one third of the coins are gold, one fifth of them are silver, two sevenths are bronze and the rest are copper. My bag can hold a maximum of 200 coins. How many coins are in my bag? A B C D 195 Solution: B (LCM of 3, 5, 7) 7 [Edexcel IGCSE(9-1) Jan F Q20b, Jan H Q6b] π΄= 2 3 Γ5Γ 7 2 Γ11 π΅= 2 4 Γ7Γ11 πΆ=3Γ 5 2 Find the lowest common multiple (LCM) of π΄, π΅ and πΆ. π π Γ π π Γ π π Γππ=ππππππ [Edexcel GCSE June2008-4H Q16b] James thinks of two numbers. He says βThe Highest Common Factor (HCF) of my two numbers is 3 The Lowest Common Multiple (LCM) of my two numbers is 45β. Write down two numbers that James could be thinking of and or 9 and 15 [Edexcel GCSE(9-1) June H Q10] Here are three lamps. Lamp A flashes every 20 seconds. Lamp B flashes every 45 seconds. Lamp C flashes every 120 seconds. The three lamps start flashing at the same time. How many times in one hour will the three lamps flash at the same time? LCM = 360 ο 10 times 10 ? 8 ? 11 ? 9 ? 12 ? ?
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Exercise 2 ? ? ? ? (on printed worksheet) N
π π finds the number of integers between 1 and π that share no factors with π other than 1 (i.e. the HCF is 1). For example π 6 =2 because for two numbers up to 6, 1 and 5, HCF(6,1) = 1 and HCF(6, 5) = 1. What is π(10)? 4 What is π(7)? What in general is π(π) for a prime number π? πβπ Given that π ππ =π π π π (provided that π»πΆπΉ π,π =1), find π =π πΓπΓπΓππ =π π π π π π π ππ =πΓπΓπΓππ=πππ Fro Note: π(π)βs posh name is βEulerβs Totient Functionβ. ? ? ? ?
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Extension :: Using Prime Factorisations
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TRUE OF FALSE? (use front of planner for true, back for false)
Using prime factorisations TRUE OF FALSE? (use front of planner for true, back for false) Q1 2Γ3 a factor of 2Γ3Γ5? True False Remarks: You can think of 2Γ3Γ5 as numbers in a Venn Diagram. If we make any selection from the numbers in this collection, we can form a factor. 3 2 5
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True False TRUE OF FALSE? Using prime factorisations
Q2 3Γ7Γ7 a multiple of 7Γ7? True False Remarks: If 3Γ7Γ7 is a multiple of 7Γ7, then 7Γ7 must be a factor of 3Γ7Γ7. 7 3 7
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True False TRUE OF FALSE? Using prime factorisations
Q3 2 3 a factor of 2 5 ? True False 2 2 Remarks: Yes, if we have 2Γ2Γ2Γ2Γ2, we can select three of these 2s to form a factor. More generally, 2 π will be a factor of 2 π provided that πβ€π. 2 2 2
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True False TRUE OF FALSE? Using prime factorisations
Q4 2 2 Γ3 a factor of 2Γ 3 3 ? True False 3 Remarks: No. We have one 2 and three 3s available Γ3 canβt be a factor because we donβt have two 2s available. 3 2 3
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True False TRUE OF FALSE? Using prime factorisations
Q5 Is 15 a factor of 2 3 Γ 3 2 Γ5? True False 3 2 2 Remarks: Yes, we have three 2s, two 3s and one 5 available. We can select a 3 and a 5 to form a factor of 15. 3 2 5
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True False TRUE OF FALSE? Using prime factorisations
Q6 2 5 Γ 3 2 Γ5 a multiple of 2 2 Γ 3 3 ? True False 2 2 Remarks: We could equivalently ask whether 2 2 Γ 3 3 is a factor of 2 5 Γ 3 2 Γ5. The answer is no because there are only two 3s available in 2 5 Γ 3 2 Γ5, so a factor couldnβt use three 3s. 2 3 5 2 3 2
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Listing factors TASK: List all the factors of 2 2 Γ 3 2
(leaving each factor in prime factorised form) Help: Remember we have a factor of a number if we make some selection from all the prime factors. 2 2 Γ Γ Γ Γ ? ? ? 2 3 ? ? 2 3 ? ? ? We get 1 as the factor in the special case where we use none of the prime factors (note that 1 is not prime!) ?
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Further Practice 1 List all the factors of 3 3 Γ5 (leaving each factor in prime factorised form) 2 List all the factors of (leaving each factor in prime factorised form) ? 3 3 Γ Γ Γ ?
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How many factors does π π Γ π π have?
Number of factors How many factors does π π Γ π π have? We can see that 2 2 Γ 3 2 has 9 factors. But could we have obtained this number without having to list them all out? 2 2 Γ Γ Γ Γ Q: How many possibilities were there for the number of 2s we use for a particular factor? 3 possibilities: We use 2 of them, 1 of them, or none at all. ? Q: How many possibilities were there for the number of 3s we use for a particular factor? Again 3 possibilities ? Q: Therefore how many total possibilities (i.e. factors) are there? For each of the 3 possible number of 2s, there are 3 possible number of 3s. πΓπ=π which is what we expected! ?
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How many factors does 3 5 Γ 5 6 have?
Number of factors How many factors does 3 5 Γ 5 6 have? There are 6 possibilities for number of 3s chosen for factor (none, 1, 2, 3, 4, 5). There are 7 possibilities for number of 5s chosen for factor (none, 1, 2, 3, 4, 5, 6). Num factors =πΓπ=ππ ? ! To get the number of factors of a number in prime factorised form, add one to each power and times the powers together. Quickfire Questions How many factors does each of the following have? 2 2 Γ β πΓπ=ππ factors 3 4 Γ5 β πΓπ=ππ factors 2Γ β πΓπ=ππ factors β ππ factors β ππ factors 3 2 Γ5Γ 7 4 β πΓπΓπ=ππ factors 5Γ7Γ11 β πΓπΓπ=π factors ? 10 3 = π π Γ π π β πΓπ=ππ factors = π π Γ π π βπΓπ=ππ factors ? ? ? ? ? ? ? ?
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Test Your Understanding
1 How many factors does 5 3 Γ 7 4 have? ? 4Γ5=20 factors 2 By first finding the prime factorisation, find the number of factors of 120. ? 120= 2 3 Γ3Γ5 β΄4Γ2Γ2=16 factors 3 Is 42 a factor of 2 2 Γ5Γ 7 3 ? ? No, as 42=2Γ3Γ7, but 2 2 Γ5Γ 7 3 does not have a 3. 4 How many factors does have? ? = Γ = Γ 5 100 β΄101Γ101=10201 factors.
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Exercise 3 Without listing out the factors, work out how many factors each of the following numbers have: 3 2 Γ factors 5 2 Γ7 6 factors 2Γ3Γ5Γ factors factors 2 10 Γ factors Work out how many factors each of the following numbers have: = π π Γ π π βππ factors = π π Γ π π βππ factors = π ππ Γ π ππ βπππ factors = π ππ Γ π ππ βπππ factors [JMC 2000 Q23] A certain number has exactly eight factors including 1 and itself. Two of its factors are 21 and 35. What is the number? Note that ππ=πΓπ and ππ=πΓπ. So the number must be some multiple of πΓπΓπ=πππ. But this does have 8 factors, as πΓπΓπ=π. Which of the following are factors of Γ 7 2 Γ11? Answer true of false. True False 5Γ True 7 2 Γ False 7Γ False List all of the twelve factors of 7 3 Γ 5 2 , leaving your factors in prime factorised form. π π Γ π π , π π Γπ, π π π π Γ π π , π π Γπ, π π πΓ π π , πΓπ, π π π , π, π Answer true or false. 3 4 is a factor of True 3 4 Γ5 is a multiple of 3 2 Γ False 35 is a factor of 2Γ 3 3 Γ False 24 is a factor of 2 4 Γ 3 2 Γ True, as ππ= π π Γπ; enough 2s and 3s are available within π π Γ π π Γπ. 1 4 ? ? ? ? ? ? ? ? ? ? 2 5 ? ? ? ? ? 6 3 ? ? ? ? ?
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Exercise 3 [JMO 1997 B2] Every prime number has two factors. How many integers between 1 and 200 have exactly four factors? Solution: 59. To have four factors the number has to be of the form π π , or πΓπ, where π and π are prime numbers. N ?
![Exercise 3 [JMO 1997 B2] Every prime number has two factors. How many integers between 1 and 200 have exactly four factors](http://slideplayer.com/slide/17887315/106/images/35/Exercise+3+%5BJMO+1997+B2%5D+Every+prime+number+has+two+factors.+How+many+integers+between+1+and+200+have+exactly+four+factors.jpg "Solution: 59. To have four factors the number has to be of the form π π , or πΓπ, where π and π are prime numbers. N.")
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