Module 11 of 45Lesson 1 of 12 · Unit 9 · Building Blocks
Number · Unit N09
Building Blocks
Every whole number is built out of primes, and there is only one way to build it. Find those building blocks and a whole family of questions collapses into the same short piece of work: how many identical hampers can be packed from 96, 72 and 120 items, when two combis next leave the rank together, and whether 8316 divides exactly by 12.
By the end of this unit you can
- Use the tests for 2, 3, 4, 5, 6, 8, 9 and 10 to decide divisibility without dividing
- List every factor of a number in pairs, and explain why 1 is not a prime number
- Break a number into primes with a factor tree and write the answer in index form
- Read the HCF of two numbers from a Venn diagram of their prime factors
- Read the LCM from the same diagram, and check it with HCF × LCM = the product
- Decide from the wording of a problem whether it wants the HCF or the LCM, and defend the choice with a size test
The one idea behind all of it
Every whole number greater than 1 is a product of prime numbers, and there is exactly one way to write it. That one fact runs the whole unit. The prime factorisation tells you what divides the number, how many factors it has, what it shares with another number — the HCF — and what the two of them build together — the LCM.
360 = 2³ × 3² × 5
One number, one factorisation — and every answer in this unit is read off it.
Two numbers, side by side, give you two more answers: what they share is the highest common factor, and what they build together is the lowest common multiple. One Venn diagram holds both.
Give every final answer to 3 significant figures, unless the answer is exact. An exact answer such as 25, 0.5 or 4 000 is written as it is, and the rounding is done once, at the very end.
Where this unit sits
This unit is about whole numbers only. It picks up where N01 Know Your Numbers left off: there you learned to calculate with integers, here you take them apart. Nothing in this unit needs a calculator, and nothing in it is a fraction.
It bridges forward to N10 Squares, Roots & Powers, where the index notation you meet here — the small raised number in 2³ — becomes the subject in its own right. The full laws of indices belong to the Algebra course, unit 11 The Power Rules; the two courses share the notation and never repeat each other’s work.
You have already used both of this unit’s big ideas without naming them. Simplifying a fraction is dividing top and bottom by their HCF, and adding fractions with different denominators means finding their LCM — both from N02 Fraction Fluency. The two-circle Venn diagram drawn here is the same diagram the Geometry strand uses for sets.
1
Tests that save you time
Divisibility by 2, 3, 4, 5, 6, 8, 9 and 10 — without dividing
Before you divide anything, look at it. A handful of tests decide whether a number is divisible by 2, 3, 4, 5, 6, 8, 9 or 10 in a couple of seconds, and they are what get a factor tree started. Two of them do nearly all the work: is it even, and does the digit sum go into 3?
| Divisible by | The test | 5148 |
|---|---|---|
| 2 | the last digit is even (0, 2, 4, 6 or 8) | yes |
| 3 | the digit sum is a multiple of 3 | yes |
| 4 | the last two digits form a multiple of 4 | yes |
| 5 | the last digit is 0 or 5 | no |
| 6 | it passes the test for 2 AND the test for 3 | yes |
| 8 | the last three digits form a multiple of 8 | no |
| 9 | the digit sum is a multiple of 9 | yes |
| 10 | the last digit is 0 | no |
Worked example 1 · Test 5148 against all eight tests
1
Divisible by 2? The last digit is 8, which is even.
yes
2
Divisible by 3? The digit sum is 5 + 1 + 4 + 8 = 18, and 18 = 3 × 6.
yes
3
Divisible by 4? The last two digits are 48, and 48 = 4 × 12.
yes
4
Divisible by 5 or by 10? The last digit is 8, not 0 and not 5.
no to both
5
Divisible by 6? It passed the test for 2 and the test for 3.
yes
6
Divisible by 8? The last three digits are 148, and 148 ÷ 8 = 18.5.
no
7
Divisible by 9? The digit sum is 18, and 18 = 9 × 2.
yes
Every one of those answers is hiding inside the prime factorisation: 5148 = 2² × 3² × 11 × 13. There are two 2s, so 4 divides it but 8 does not. There are two 3s, so 9 divides it. There is no 5, so neither 5 nor 10 divides it. The tests are quick; the prime factorisation is the reason they work.
EXAMPLEThere is no test for 7 that is quicker than dividing — so divide. And be careful with combinations: a number divisible by 4 and by 6 need not be divisible by 24, because 4 and 6 both contain a 2. 12 passes both tests and is not a multiple of 24. Combining tests only works when the two numbers share no prime factor, which is why the 12 test is “divisible by 3 and by 4” and not “by 2 and by 6”.
Practice · Tests that save you time · 10 questions
P1Write down whether 4218 is divisible by 2.Fluency
Answeryes
P2Write down whether 5431 is divisible by 3.Fluency
Answerno
P3Write down whether 7320 is divisible by 5 and by 10.Fluency
Answeryes to both
P4Write down whether 9144 is divisible by 4.Fluency
Answeryes
P5Determine whether 3852 is divisible by 6.Applied
Answeryes6 = 2 × 3, so run both tests. Last digit 2 is even, so it passes the test for 2. Digit sum 3 + 8 + 5 + 2 = 18, and 18 = 3 × 6, so it passes the test for 3. It passes both, so it is divisible by 6. Check: 3852 ÷ 6 = 642.
P6Determine whether 5184 is divisible by 8.Applied
AnsweryesTest for 8: look at the last three digits, 184. Since 184 = 8 × 23, the whole number is divisible by 8. Check: 5184 ÷ 8 = 648.
P7Determine whether 62613 is divisible by 9.Applied
AnsweryesDigit sum 6 + 2 + 6 + 1 + 3 = 18, and 18 = 9 × 2, so it is divisible by 9. Note it is not divisible by 2, because the last digit is odd. Check: 62613 ÷ 9 = 6957.
P8In the number 47d2 the digit d is missing. Determine the smallest digit d that makes the number divisible by 3.Applied
Answer2The digit sum is 4 + 7 + d + 2 = 13 + d. For divisibility by 3 the digit sum must be a multiple of 3, so 13 + d = 15, 18 or 21, giving d = 2, 5 or 8. The smallest is d = 2, and 4722 ÷ 3 = 1574.
P9A number is divisible by 12 exactly when it is divisible by both 4 and 3. Determine whether 8316 is divisible by 12, showing both tests.Challenge
AnsweryesTest for 4: the last two digits are 16, and 16 = 4 × 4, so it passes. Test for 3: the digit sum is 8 + 3 + 1 + 6 = 18, a multiple of 3, so it passes. It passes both, so it is divisible by 12. Check: 8316 ÷ 12 = 693. The pair 4 and 3 works because 4 and 3 share no prime factor.
P10Determine a number that is divisible by both 4 and 6 but is not divisible by 24, and describe why such a number exists.Challenge
Answer1212 is divisible by 4 and by 6, but 12 ÷ 24 is not a whole number. The reason is that 4 and 6 share the prime factor 2: 4 = 2² and 6 = 2 × 3, so the smallest number divisible by both is the LCM, 2² × 3 = 12, not 4 × 6 = 24. Multiplying the two tests together only works when the two numbers share no prime factor.
2
Factors, multiples and primes
Listing every factor in pairs, and what makes a number prime
A factor of a number divides into it exactly. A multiple is what you get when you multiply it. Factors come in pairs, so the safe way to list them is to walk up from 1 writing both members of each pair, and to stop when the pairs start repeating. A prime number is one with exactly two different factors — which is why 1, with its single factor, is not prime.
Prime
- Exactly two different factors: 1 and itself.
- 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, …
- 2 is prime — and it is the only even prime.
- To test n, try the primes up to √n. If none divides it, it is prime.
Not prime
- 1 has only one factor, so it is neither prime nor composite.
- 51 = 3 × 17 and 91 = 7 × 13 — odd is not the same as prime.
- Every even number except 2 has 2 as a third factor.
- A composite number is one with more than two factors.
Worked example 2 · Every factor of 48, and is 51 prime?
1
Start at 1 and work upwards in pairs. 1 × 48, 2 × 24, 3 × 16, 4 × 12.
four pairs so far
2
5 is not a factor (48 does not end in 0 or 5). 6 × 8 is the next pair.
1, 2, 3, 4, 6, 8, 12, 16, 24, 48
3
Stop at 7, because 7 × 7 = 49 is already past 48 — every pair has been found.
10 factors
4
Now 51: it is odd, and its digit sum is 5 + 1 = 6, a multiple of 3.
3 divides it
5
51 = 3 × 17, so 51 has the factors 1, 3, 17 and 51.
51 is not prime
Factors arrive in pairs, which is why most numbers have an even number of them. A square number is the exception: 36 = 6 × 6 uses the same factor twice, so 36 has nine factors, an odd number. Hold on to that — the Criterion B investigation below is built on it.
Practice · Factors, multiples and primes · 10 questions
P11Write down all the factors of 18.Fluency
Answer1, 2, 3, 6, 9, 18
P12Write down the first four multiples of 7.Fluency
Answer7, 14, 21, 28
P13Write down whether 1 is a prime number.Fluency
Answerno
P14Write down the prime numbers between 20 and 30.Fluency
Answer23, 29
P15Determine all the factors of 48 by working through the factor pairs.Applied
Answer1, 2, 3, 4, 6, 8, 12, 16, 24, 48Work upwards in pairs: 1 × 48, 2 × 24, 3 × 16, 4 × 12, 6 × 8. 5 is not a factor. Stop at 7, because 7 × 7 = 49 is already past 48, so every pair has been found. That is 10 factors.
P16Determine whether 91 is a prime number.Applied
Answerno, 91 = 7 × 13Test the primes in turn: 91 is odd, its digit sum is 10 so 3 fails, it does not end in 0 or 5, and 91 ÷ 7 = 13 exactly. Since 91 has a third factor it is not prime. You only ever have to test primes up to the square root: 9² = 81 and 10² = 100, so testing 2, 3, 5 and 7 was enough.
P17Determine how many factors 36 has.Applied
Answer9Pair them off: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6. The last pair is a repeat, so 6 is counted once: 1, 2, 3, 4, 6, 9, 12, 18, 36 — nine factors.
P18Determine the largest prime number below 100.Applied
Answer97Work down from 99: 99 has digit sum 18, so 3 divides it; 98 is even; 97 is odd, has digit sum 16, does not end in 0 or 5, and 97 ÷ 7 = 13.857…. Since 10² = 100 > 97, testing 2, 3, 5 and 7 is enough, so 97 is prime.
P19Determine every number below 50 that has exactly three factors, and describe what they have in common.Challenge
Answer4, 9, 25 and 494 has factors 1, 2, 4; 9 has 1, 3, 9; 25 has 1, 5, 25; 49 has 1, 7, 49. Each one is a prime squared — 2², 3², 5², 7². A number p² has exactly the three factors 1, p and p², and no other kind of number can have exactly three.
P20A number has exactly two factors and is even. Determine the number.Challenge
Answer2Exactly two factors means the number is prime. Every even number bigger than 2 has 2 as a third factor, so it cannot be prime. That leaves 2 itself, whose only factors are 1 and 2. 2 is the only even prime number.
3
Breaking a number into primes
The factor tree, and writing the answer in index form
A factor tree takes a number apart until every branch ends in a prime. Split it any way you can see — the tree looks different each time and the circled primes never change. Collect equal primes into powers and you have the index form, which is the form every later question in this unit wants.
Split 360 any way you like: the branches differ, the six circled primes never do. Primes are shown in the filled circles — a branch stops only when it reaches one.
Worked example 3 · Factor tree of 360, and the index form
1
Choose any split you can see. 360 = 36 × 10.
two branches
2
Neither is prime, so split again: 36 = 6 × 6 and 10 = 2 × 5.
2 and 5 are prime
3
6 is still not prime: 6 = 2 × 3, twice over.
every branch now ends
4
Circle the leaves and write them in order.
2 × 2 × 2 × 3 × 3 × 5
5
Collect equal primes into powers.
360 = 2³ × 3² × 5
6
Check by multiplying back: 8 × 9 × 5.
360 ✓
The first split does not matter. Starting 360 = 4 × 90 or 360 = 2 × 180 gives a different-looking tree and exactly the same six primes. Every whole number bigger than 1 has one and only one prime factorisation — that is why it is worth finding.
EXAMPLEIndex form is not decoration, it is compression. 2⁵ × 3 says everything about 96 in five characters: it is even five times over, so 2, 4, 8, 16 and 32 all divide it; 3 divides it; 5 does not; and it has (5 + 1) × (1 + 1) = 12 factors — a count you will prove for yourself in the Criterion B task.
Practice · Breaking a number into primes · 10 questions
P21Write down 24 as a product of primes in index form.Fluency
Answer2³ × 3
P22Write down 45 as a product of primes in index form.Fluency
Answer3² × 5
P23Write down the value of 2² × 5³.Fluency
Answer500
P24Write down the prime factorisation of 100 in index form.Fluency
Answer2² × 5²
P25Determine the prime factorisation of 360 using a factor tree.Applied
Answer2³ × 3² × 5Split it any way you like and keep going until every branch ends in a prime: 360 = 36 × 10, then 36 = 6 × 6 and 10 = 2 × 5, then each 6 = 2 × 3. The circled primes are 2, 3, 2, 3, 2, 5. In order: 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5.
P26Determine the prime factorisation of 588.Applied
Answer2² × 3 × 7²588 is even: 588 = 2 × 294, and 294 = 2 × 147. Now 147 has digit sum 12, so 147 = 3 × 49, and 49 = 7 × 7. Altogether 588 = 2 × 2 × 3 × 7 × 7 = 2² × 3 × 7².
P27Determine the prime factorisation of 1000 in index form.Applied
Answer2³ × 5³1000 = 10 × 10 × 10 and 10 = 2 × 5, so 1000 = (2 × 5)³ = 2³ × 5³. Every power of 10 has equal numbers of 2s and 5s.
P28Determine the number whose prime factorisation is 2⁴ × 3 × 5².Applied
Answer12002⁴ = 16 and 5² = 25, so the number is 16 × 3 × 25 = 48 × 25 = 1200.
P29Determine the prime factorisation of 1752.Challenge
Answer2³ × 3 × 731752 is even three times over: 1752 = 2 × 876 = 2 × 2 × 438 = 2 × 2 × 2 × 219. Now 219 has digit sum 12, so 219 = 3 × 73, and 73 is prime (test 2, 3, 5 and 7 — none divides it, and 9² = 81 > 73). So 1752 = 2³ × 3 × 73.
P30A number has prime factorisation 2a × 3. Determine the smallest value of a for which the number is greater than 100.Challenge
Answer6Build the list: 2¹ × 3 = 6, then 12, 24, 48, 96, 192. 2⁵ × 3 = 96 is still below 100, and 2⁶ × 3 = 192 is above it, so a = 6.
4
The highest common factor
Two circles, one overlap — the HCF is what they share
The highest common factor of two numbers is the biggest number that divides into both. Break each number into primes, draw two circles, and put every prime they share in the overlap. Multiply the overlap and you have the HCF. If the overlap is empty the HCF is 1, and the two numbers are called co-prime.
Two trees, two index forms — everything the Venn diagram below needs.
The overlap is the HCF. The whole diagram, taking each circle’s leftovers once, is the LCM. One picture answers both questions.
Worked example 4 · HCF of 84 and 120 from a Venn diagram
1
Break each number into primes.
84 = 2² × 3 × 7
2
And the other one.
120 = 2³ × 3 × 5
3
Put the shared primes in the overlap: one 2, a second 2, and a 3. The 7 is left in the 84 circle; a third 2 and the 5 are left in the 120 circle.
overlap = 2, 2, 3
4
Multiply the overlap.
HCF = 2 × 2 × 3 = 12
5
Check: 84 ÷ 12 = 7 and 120 ÷ 12 = 10, and 7 and 10 share no factor.
12 cannot be beaten
The overlap is the HCF, always. If the two circles share nothing, the overlap is empty and the HCF is 1 — the numbers are then called co-prime.
| What you take | 84 = 2² × 3 × 7 and 120 = 2³ × 3 × 5 | |
|---|---|---|
| HCF | the lower power of every shared prime | 2² × 3 = 12 |
| LCM | the higher power of every prime that appears | 2³ × 3 × 5 × 7 = 840 |
Practice · The highest common factor · 10 questions
P31Write down the HCF of 8 and 12.Fluency
Answer4
P32Write down the HCF of 9 and 15.Fluency
Answer3
P33Write down the HCF of 7 and 13.Fluency
Answer1
P34Write down the HCF of 6 and 24.Fluency
Answer6
P35Determine the HCF of 84 and 120 using prime factors.Applied
Answer1284 = 2² × 3 × 7 and 120 = 2³ × 3 × 5. Take the lower power of every prime they share: 2² and 3. The HCF is 2² × 3 = 12. Check: 84 ÷ 12 = 7 and 120 ÷ 12 = 10, and 7 and 10 share nothing, so nothing bigger will work.
P36Determine the HCF of 90 and 126.Applied
Answer1890 = 2 × 3² × 5 and 126 = 2 × 3² × 7. The shared part is 2 × 3² = 18. Check: 90 ÷ 18 = 5 and 126 ÷ 18 = 7.
P37Determine the HCF of 45, 60 and 75.Applied
Answer1545 = 3² × 5, 60 = 2² × 3 × 5 and 75 = 3 × 5². Only 3 and 5 appear in all three, and the lowest power of each is 3¹ and 5¹, so the HCF is 3 × 5 = 15. There is no 2 in the HCF because 45 and 75 are odd.
P38Two numbers have prime factorisations 2³ × 3² × 5 and 2² × 3 × 7. Determine their HCF, leaving your answer in index form.Applied
Answer2² × 3, which is 12Compare prime by prime and take the lower power each time: for 2 that is 2², for 3 it is 3¹, and 5 and 7 are missing from one number each so they are not in the HCF. The HCF is 2² × 3 = 12. You never need the numbers themselves — 360 and 84 — to do this.
P39Determine the highest common factor of 1752 and 356.Challenge
Answer41752 = 2³ × 3 × 73 and 356 = 2² × 89. The only shared prime is 2, and the lower power is 2², so the HCF is 4. Check: 1752 ÷ 4 = 438 and 356 ÷ 4 = 89, and 89 is prime and is not a factor of 438, so 4 cannot be beaten.
P40The HCF of two numbers is 6, and one of the numbers is 24. Determine the smallest the other number could be.Challenge
Answer6The other number must be a multiple of 6, so the candidates are 6, 12, 18, 24, …. HCF(24, 6) = 6, so 6 works and nothing smaller is a multiple of 6. Note 12 would give HCF(24, 12) = 12, which is too big, so bigger candidates are not automatically better.
5
The lowest common multiple
The same diagram, read the other way — everything, once each
The lowest common multiple of two numbers is the smallest number they both divide into. It is the same diagram, read whole: the overlap once, plus whatever is left in each circle. And there is a free check you should always use — HCF × LCM is the product of the two numbers.
Worked example 5 · LCM of 84 and 120 from the same diagram
1
Do not redraw anything. Read the whole diagram this time.
overlap 2, 2, 3 · left 7 · right 2, 5
2
Multiply everything in the diagram, taking each circle’s leftovers once.
LCM = 12 × 7 × 2 × 5
3
Work it out.
LCM = 840
4
Check both ways round: 840 ÷ 84 and 840 ÷ 120.
10 and 7 ✓
5
Check the pairing rule: HCF × LCM = 12 × 840, and 84 × 120.
10080 = 10080 ✓
One diagram, two answers. The overlap alone is the HCF; the whole diagram is the LCM. And HCF × LCM = the product of the two numbers, every time — a free check you should use whenever you have found both.
EXAMPLEThe check catches the commonest slip of all. If you multiply 84 × 120 and call it the LCM you get 10080, but 10080 ÷ 12 = 840, so 10080 counts the shared 12 twice. Multiplying the two numbers together always gives a common multiple — it is the lowest one only when the numbers share nothing, as with 5 and 8.
Practice · The lowest common multiple · 10 questions
P41Write down the LCM of 4 and 6.Fluency
Answer12
P42Write down the LCM of 5 and 8.Fluency
Answer40
P43Write down the LCM of 3 and 12.Fluency
Answer12
P44Write down the LCM of 10 and 15.Fluency
Answer30
P45Determine the LCM of 84 and 120 using prime factors.Applied
Answer84084 = 2² × 3 × 7 and 120 = 2³ × 3 × 5. Take the higher power of every prime that appears: 2³, 3, 5 and 7. The LCM is 8 × 3 × 5 × 7 = 840. Check: 840 ÷ 84 = 10 and 840 ÷ 120 = 7.
P46Determine the LCM of 12 and 18.Applied
Answer3612 = 2² × 3 and 18 = 2 × 3². Higher power of each prime: 2² and 3². The LCM is 4 × 9 = 36. Check: 36 ÷ 12 = 3 and 36 ÷ 18 = 2.
P47Two numbers have prime factorisations 2³ × 3 and 2 × 3² × 5. Determine their LCM.Applied
Answer2³ × 3² × 5, which is 360Take the higher power of each prime: 2³ from the first, 3² and 5 from the second. The LCM is 8 × 9 × 5 = 360. The two numbers are 24 and 90, and 360 ÷ 24 = 15 while 360 ÷ 90 = 4.
P48Determine the LCM of 8, 12 and 20.Applied
Answer1208 = 2³, 12 = 2² × 3 and 20 = 2² × 5. The highest power of 2 anywhere is 2³, and 3 and 5 each appear once, so the LCM is 8 × 3 × 5 = 120. Check: 120 ÷ 8 = 15, 120 ÷ 12 = 10 and 120 ÷ 20 = 6.
P49Two numbers have HCF 12 and LCM 840. One of the numbers is 84. Determine the other number.Challenge
Answer120For any two numbers, HCF × LCM = the product of the numbers. So 12 × 840 = 84 × the other number, giving 10080 = 84 × n and n = 10080 ÷ 84 = 120. Check: HCF(84, 120) = 12 and LCM(84, 120) = 840.
P50Determine the smallest number greater than 1 that leaves a remainder of 1 when it is divided by 4, by 5 and by 6.Challenge
Answer61A number that leaves remainder 1 is one more than a common multiple. The smallest common multiple of 4, 5 and 6 is the LCM: 4 = 2², 5 = 5, 6 = 2 × 3, so the LCM is 2² × 3 × 5 = 60. The number is 60 + 1 = 61. Check: 61 = 4 × 15 + 1 = 5 × 12 + 1 = 6 × 10 + 1.
6
Which one does the question want?
Packing, tiling and syncing — choosing HCF or LCM and defending it
In a word problem nobody tells you which one to use. Sort it before you calculate: are you cutting something up or bringing things back together? Then check the size of your answer against the numbers in the question. That one habit turns the most-missed marks in this unit into free ones.
| The question sounds like | You need | Size check |
|---|---|---|
| equal bags, equal groups, sharing out, the largest tile, cutting into equal pieces, the biggest spacing | HCF | never bigger than the smallest number given |
| flashing together, meeting again, the next time both, laps of a track, the smallest number both divide into | LCM | never smaller than the biggest number given |
The lights flash together every 36 seconds — the LCM. The HCF of 6 seconds is also on the diagram, but it answers a different question.
Worked example 6 · Two lights, 12 seconds and 18 seconds
1
Read the question: they start together and we want the NEXT time together, so the answer must be bigger than both numbers.
this is an LCM
2
Break both into primes.
12 = 2² × 3 · 18 = 2 × 3²
3
Take the higher power of each prime.
2² × 3² = 36
4
Check by listing: 12, 24, 36 and 18, 36.
36 s is the first match
5
In a 3-minute advert, 180 ÷ 36.
5 times together
The words to hunt for are “together again”, “next time” and “smallest number that both divide”. All three mean LCM, and all three have an answer at least as big as the largest number in the question.
Here it is the overlap that matters: 24 identical bags, each holding 96 ÷ 24 = 4 sweets and 72 ÷ 24 = 3 biscuits.
Worked example 7 · Equal gift bags from 96 sweets and 72 biscuits
1
Read the question: the bags are cut out of what is already there, so the answer cannot be bigger than 72.
this is an HCF
2
Break both into primes.
96 = 2⁵ × 3 · 72 = 2³ × 3²
3
Take the lower power of each shared prime.
2³ × 3 = 24
4
Share out what each bag gets: 96 ÷ 24 and 72 ÷ 24.
4 sweets, 3 biscuits
5
Check nothing is left over: 24 × 4 and 24 × 3.
96 and 72 ✓
Same two numbers, opposite question. The LCM of 96 and 72 is 288, which would be more bags than you have sweets — an answer that fails the common-sense test before you even check the arithmetic.
EXAMPLEA combi leaves for Mogoditshane every 15 minutes and one leaves for Tlokweng every 24 minutes, both starting at 06:00. LCM(15, 24) = 2³ × 3 × 5 = 120 minutes, so they next leave together at 08:00. Worth saying out loud: at a real rank a combi leaves when it is full, so the answer is what the timetable promises, not what the rank delivers.
Practice · Which one does the question want? · 10 questions
P51Two combis leave the rank at the same moment and then run on fixed intervals. Write down whether you need the HCF or the LCM to find when they next leave together.Fluency
Answerthe LCM
P52Two ribbons are to be cut into equal pieces that are as long as possible, with nothing left over. Write down whether you need the HCF or the LCM.Fluency
Answerthe HCF
P53Two lights flash every 4 seconds and every 6 seconds. They flash together now. Write down after how many seconds they next flash together.Fluency
Answer12 seconds
P54Write down the largest number of identical party bags that can be made from 12 pens and 18 notebooks with nothing left over.Fluency
Answer6
P55One light flashes every 12 seconds and another every 18 seconds. They flash together at 07:00. Determine after how many seconds they next flash together.Applied
Answer36 secondsThis is a “when do they meet again” question, so it is the LCM. 12 = 2² × 3 and 18 = 2 × 3², so the LCM is 2² × 3² = 36. The first light flashes at 12, 24, 36 s and the second at 18, 36 s — 36 s is the first time both appear.
P56A shop has 96 sweets and 72 biscuits. Identical bags are made using all of them. Determine the largest number of bags, and what goes into each bag.Applied
Answer24 bags, each with 4 sweets and 3 biscuitsThe bags are cut out of what is already there, so it is the HCF. 96 = 2⁵ × 3 and 72 = 2³ × 3², so the HCF is 2³ × 3 = 24. Each bag then holds 96 ÷ 24 = 4 sweets and 72 ÷ 24 = 3 biscuits. Check: 24 × 4 = 96 and 24 × 3 = 72.
P57Combis on one route leave a rank every 15 minutes and on another route every 24 minutes. Both leave at 06:00. Determine the next time both routes leave together.Applied
Answer08:00It is the LCM. 15 = 3 × 5 and 24 = 2³ × 3, so the LCM is 2³ × 3 × 5 = 120 minutes, which is 2 hours. 06:00 + 2 hours = 08:00.
P58A floor 96 cm by 72 cm is to be covered exactly by identical square tiles, with no tile cut. Determine the largest tile that can be used and how many tiles are needed.Applied
Answer24 cm tiles, and 12 of themThe tile has to divide both 96 and 72, so the largest is the HCF, 24 cm. The floor is then 96 ÷ 24 = 4 tiles one way and 72 ÷ 24 = 3 tiles the other, so 4 × 3 = 12 tiles. Check: 12 × 24² = 6912 cm², and 96 × 72 = 6912 cm².
P59Three routes leave a rank every 15, 20 and 24 minutes, all together at 06:00. Determine the next time all three leave together, and describe one reason the real rank might not match your answer.Challenge
Answer08:0015 = 3 × 5, 20 = 2² × 5 and 24 = 2³ × 3. The LCM takes the highest power of each: 2³ × 3 × 5 = 120 minutes = 2 hours, so 08:00. In a real rank a combi usually leaves when it is full rather than on a fixed interval, so the model is an idealisation and the true times will drift.
P60A rectangular plot is 132 m by 84 m. Fence posts are placed all the way round, equally spaced, with a post at each corner. Determine the largest possible spacing and the number of posts needed.Challenge
Answer12 m apart, and 36 postsFor a post to land on every corner the spacing must divide both 132 and 84, so the largest is the HCF. 132 = 2² × 3 × 11 and 84 = 2² × 3 × 7, so the HCF is 2² × 3 = 12 m. The perimeter is 2 × (132 + 84) = 432 m, and posts every 12 m round a closed loop need 432 ÷ 12 = 36 posts (going right round, the last post is the first one again, so there is no +1 here).
The building-blocks routine
Four moves, whatever the question.
1
Run the quick tests firstIs it even? Does the digit sum go into 3? Does it end in 0 or 5? Three glances usually give you the first split of the tree, and they cost nothing.
2
Build the tree and circle every primeAny first split works. Keep splitting until every branch ends in a circled prime — a leaf you have not circled is a branch you have not finished.
3
Write it in index form, primes in order2 × 2 × 2 × 3 × 3 × 5 becomes 2³ × 3² × 5. Now the number is in the form every later question wants.
✓
Ask which answer the question wantsCutting up, packing, sharing, largest tile → HCF, and the answer is never bigger than the smallest number given. Meeting again, flashing together, smallest number both divide → LCM, and the answer is never smaller than the biggest number given.
The mistakes that cost the most marks
11 is counted as a prime number
Wrong“The factors of 12 are 1, 2, 3, 4, 6 and 12, so its prime factors are 1, 2 and 3, and 12 = 1 × 2² × 3.”
RightA prime number has exactly two different factors. 1 has only one factor — itself — so 1 is not prime (and it is not composite either). 12 = 2² × 3.
If 1 were allowed, 12 could also be written 1 × 2² × 3, or 1² × 2² × 3, or 1⁹⁹ × 2² × 3, and a number would no longer have exactly one prime factorisation. That is the real reason 1 is excluded. While you are here: 2 is prime — it is the only even prime, because every other even number has 2 as a third factor.
2HCF and LCM swapped in a word problem
Wrong“96 sweets and 72 biscuits, so the number of bags is the LCM, 288.”
RightThe bags are made out of what you already have, so the answer must be smaller than 72. It is the HCF: 24 bags.
Sort the question before you calculate. Cutting up, sharing out, packing into equal groups, the largest tile → HCF, and the answer is never bigger than the smallest number in the question. Meeting again, flashing together, the smallest number both divide into → LCM, and the answer is never smaller than the biggest number in the question. Check your answer against that size test every time.
3The factor tree is stopped before every leaf is prime
Wrong“360 = 8 × 45 = 8 × 9 × 5, so the prime factorisation is 8 × 9 × 5.”
Right8 and 9 are not prime, so the tree is not finished. 8 = 2 × 2 × 2 and 9 = 3 × 3, giving 360 = 2³ × 3² × 5.
Before you stop, run each leaf past the tests: is it even? is its digit sum a multiple of 3? does it end in 0 or 5? A leaf that answers yes to any of those can still be split. Circle every prime as you finish it — an uncircled leaf is an unfinished branch.
The two pale leaves are still composite: 8 = 2 × 2 × 2 and 9 = 3 × 3. Finish them and 360 = 2³ × 3² × 5.
Tasks to submit
Criterion B · Investigating patterns
How many factors does a number have?
Counting the factors of a number by listing them is slow, and it is easy to miss one. There is a much faster way, and it is hiding in the index form. This investigation asks you to find it.
Two 2s and two 3s. The Criterion B investigation asks you to find what the indices 2 and 2 have to do with the nine factors of 36.
Part 1 · structured
| n | Every factor of n | How many factors | Prime factorisation |
|---|---|---|---|
| 6 | 1, 2, 3, 6 | 4 | 2 × 3 |
| 8 | 1, 2, 4, 8 | 4 | 2³ |
| 12 | |||
| 16 | |||
| 18 | |||
| 20 | |||
| 24 | |||
| 36 |
- Complete the table. Take your time over the factor lists — the whole investigation depends on them being right, so pair them off as in worked example 2.
- Predict how many factors 48 has, and how many 100 has, before you list either of them. Write both predictions down first, then check them by listing.
- Identify what is going on. Compare the indices in the last column with the number in the third column, row by row. Say in words what you notice, first for the numbers built from one prime and then for the numbers built from two.
- Determine a general rule for the number of factors of n = pa × qb, where p and q are different primes. Your rule must work for any a and any b, and it must say what to do when there is only one prime.
- Test and verify your rule on 84 and on 200, by using the rule and then by listing every factor the long way. The two answers must agree, and you should show both.
- Justify why the rule works. Think about how a factor of n is built out of the primes of n: how many 2s could it take from 2³, and how many 3s from 3²? Explain why that counting gives your rule, and why the “+ 1” is there.
Part 2 · open
Now you choose what to investigate, and how. Here are three openings — take one, or propose your own. You decide which numbers to try, how many you need, how to record them and how you will test whatever you find.
- Which numbers have exactly three factors? Investigate, find them all below 200, and use your rule from Part 1 to say exactly what kind of number they are — and why there can be no others.
- Which numbers have an odd number of factors? Investigate. The answer is a famous family of numbers, and worked example 2 has already dropped a hint. Explain what makes them different, in terms of factor pairs.
- Which number below 100 has the most factors? Investigate, and then determine whether the biggest numbers are the ones with the most factors — or whether something else decides it.
Whichever you choose, take it through the same ladder: predict, identify the pattern, determine a general rule, test and verify it on cases you have not already used, and justify why it must hold and where it stops holding.
Submit: your completed table, your rule and your reasoning, as a scan or a typed document. Marked out of 8 against Criterion B. Completing the table earns the lower bands; the general rule, the verification and the justification are what reach levels 5–8. Part 2 must show your own choice of method.
Remember, you should communicate in an organized and coherent manner.
Criterion D · Applying maths in real-life contexts
Hampers, combis and a Saturday at the market
Mma Kelebogile Selepeng runs a stall at the Broadhurst market in Gaborone. On the last Saturday of the month she wants to make up identical hampers from stock she already has, and she needs to know when the two combi routes her customers use leave the rank together, so that she can be standing at the stall when they arrive. Both tables below contain more information than you need.
Source 1 · the stock list
| Item | In stock | Unit cost | Supplier | Shelf |
|---|---|---|---|---|
| Bars of soap | 96 | P6.50 | Gaborone Wholesalers | A2 |
| Packets of sugar (2 kg) | 72 | P12.95 | Gaborone Wholesalers | A4 |
| Boxes of tea (100 bags) | 120 | P8.75 | Kgale Traders | B1 |
| Tins of jam | 45 | P14.20 | Kgale Traders | B3 |
| Packets of biscuits | 17 | P9.60 | Riverwalk Suppliers | C1 |
| Bottles of cooking oil (750 ml) | 30 | P22.40 | Kgale Traders | C2 |
| Bags of maize meal (5 kg) | 54 | P39.95 | Gaborone Wholesalers | D1 |
She has decided that every hamper will contain soap, sugar and tea, and that no item of those three may be left over. She sells each hamper for P135.00. Prices are in pula and thebe; names, quantities and prices are illustrative and were written for this task.
Source 2 · the combi rank board
| Route | Where it goes | Minutes between departures | Fare | Distance |
|---|---|---|---|---|
| Route 1 | Gaborone rank → Mogoditshane | 15 | P5.50 | 12 km |
| Route 2 | Gaborone rank → Tlokweng | 24 | P6.00 | 15 km |
| Route 3 | Gaborone rank → Mochudi | 40 | P12.50 | 37 km |
| Route 4 | Gaborone rank → Gabane | 20 | P7.00 | 18 km |
| Route 5 | Gaborone rank → Ramotswa | 45 | P14.00 | 30 km |
Her customers use Route 1 and Route 2. Every route sends its first combi at 06:00 and its last at 18:00. The board assumes each route keeps to its stated interval exactly.
- Identify the information you actually need from the two tables, and describe why each column you are leaving out does not matter for these two questions. (Strand i — identifying the relevant elements)
- Select your strategies. Write down which of the two questions needs the HCF and which needs the LCM, and say how the wording told you. Then write each quantity you need as a product of primes. (Strand ii — selecting appropriate strategies)
- Apply them. Determine the largest number of identical hampers she can make, what goes into one hamper, what one hamper costs her (give money to two decimal places) and what she makes on the whole batch. Then determine the next time after 06:00 that both routes leave together, and how many times they do so before the rank closes. (Strand iii — applying the strategy)
- Describe how accurate your two answers can honestly be. The stock counts are exact whole numbers, so what does that mean for the hamper answer? The departure intervals are not: at a real rank a combi usually leaves when it is full, not when the clock says so. Describe what that does to your 08:00 answer, and describe what rounding money to two decimal places means when you multiply a single hamper’s profit by the whole batch. (Strand iv — accuracy of the solution)
- Describe whether your answers make sense at a real market stall. She also has 17 packets of biscuits. Determine what happens to the number of hampers if she insists on using all of those too, and describe why. Then describe what she should do with the stock that is left over on the shelf, and whether making the largest possible number of hampers is actually the best decision for one Saturday. (Strand v — does it make sense in context)
Submit: your prime factorisations, your two answers with working, your money to two decimal places, and short written answers for parts 1, 4 and 5. Marked out of 8 against Criterion D. Every strand is marked, and strands iv and v carry as much weight as the calculations — an answer that ignores the leftovers, or that treats a combi timetable as if it were a train timetable, will lose marks.
When you are ready
The mastery quiz is 10 questions drawn at random from a bank of 25. You need 70% to pass and you have three attempts. Work through the practice questions above first — especially Concept 3, where the tree has to reach every prime, and Concept 6, where you decide between the HCF and the LCM.
Printable version
Download Unit 9 as a PDF — the same lesson, laid out for printing and for working offline.
Download Unit 9 as a PDF — the same lesson, laid out for printing and for working offline.