Download Yr7-NumberTheory (Slides)

Year 7 Number Theory
Dr J Frost ([email protected])
www.drfrostmaths.com
Objectives: Have an appreciation of properties of integers (whole
numbers), including finding the Lowest Common Multiple, Highest
Common Factor, and using the prime factorisation of numbers for a
variety of purposes. Reason about divisibility in equations.
Last modified: 30th November 2016
For Teacher Use:
Recommended lesson structure:
Lesson 1: Introduction to Number Theory/Sums of primes + squares problems
Lesson 2: Prime Factorisation
Lesson 3: Factors using Prime Factorisation
Lesson 4: LCM/HCF
Lesson 5: Divisibility Rules
Go >
Go >
Go >
Go >
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Extension 1: Divisibility of Terms/Within Equations
Extension 2: More on Squares and Cubes
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Even More Stuff!
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Starter
Bro Pro Tip: You should
try to memorise these.
List the following numbers in your books.
The first 16 square numbers:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100,
? 225, 256
121, 144, 169, 196,
The first 8 cube numbers:
1, 8, 27, 64, 125, 216, 343, 512
The prime numbers up to 40:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,
?
37
The first 10 triangular numbers:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66
?
(e.g. 3 is a triangular number as you can form a
triangle using 1 dot on the first row and 2 on the next)
?
If you finish:
A ‘perfect’ number is a number who factors (excluding itself) add up to itself.
For example. The factors of 6 (excluding 6) are 1, 2, 3, and 1 + 2 + 3 = 6.
Find the first perfect number after 6.
Bro Fact: All perfect numbers
are triangular numbers.
Solution: 𝟐𝟖 = 𝟏 + 𝟐 +?𝟒 + 𝟕 + 𝟏𝟒
Other numbers you might want to investigate yourself:
Tetrahedric numbers, Fibonacci numbers.
Key Terms
!
!
?
Integer: A whole number.
Positive integer: An integer that
? is at least 1.
Non-negative integer: An integer that
? is at least 0.
Perfect square: For integers, just ?a square number.
Divisor: Another word
? for factor.
Composite: The opposite of prime:
? has other factors.
Distinct integers:
Numbers which ?are different!
Bro Side Note: A ‘perfect
square’ more generally refers
to ‘something squared’, which
can be an algebraic expression.
For example 𝑥 + 1 2 is a
‘perfect square’, but is not
necessarily a square number,
e.g. if 𝑥 = 0.5 (as 1.52 = 2.25)
Puzzles involving sums of primes/squares/…
Example: Goldbach’s Conjecture (as of current, unproven!) states that all
even numbers greater than 2 are the sum of two primes.
Bro Tip: It often
How many ways are there of expressing 100 as the sum of two primes?
helps to write
Solution: 6 (3 + 97, 11 + 89, 17 + 83, 29 + 71, 41 + 59, 47 + 53)
out your
?
numbers of
interest (primes,
squares, …) first.
Further Example: The Indian mathematician Ramanujan once famously noted
that the 1729 number of a taxi ridden by his friend Hardy:
“is a very interesting number; it is the smallest integer expressible as a sum
of two different cubes in two different ways”.
What is the smallest integer (not necessarily a square) that is expressible as
the sum of two distinct squares in two different ways?
(Hint: 1 is used in one of the sums)
𝟔𝟓 = 𝟒𝟗 + 𝟏𝟔 = 𝟔𝟒 + 𝟏
(Side note: the smallest square number expressible
as the sum of two
?
squares in two different ways is 𝟐𝟓𝟐 = 𝟕𝟐 + 𝟐𝟒𝟐 = 𝟐𝟎𝟐 + 𝟏𝟓𝟐)
Exercise 1
(Problems on provided sheet)
1 [JMC 2015 Q11] What is the smallest prime number that is the
sum of three different prime numbers?
A 11
B 15
C 17
D 19
E 23
Solution: D
?
2
[JMO 1999 A2] In how many different ways can 50 be written as
the sum of two prime numbers? (Note: 𝑥 + 𝑦 and 𝑦 + 𝑥 do not
count as different.)
Solution: 4 ways (𝟒𝟕 + 𝟑, 𝟒𝟑
? + 𝟕, 𝟑𝟕 + 𝟏𝟑, 𝟑𝟏 + 𝟏𝟗)
3
[JMO 2009 A3] The positive whole numbers 𝑎, 𝑏 and 𝑐 are all
different and
𝑎2 + 𝑏2 + 𝑐 2 = 121. What is the value of 𝑎 + 𝑏 + 𝑐?
Solution: 17
?
Bro Tip: Again,
use your lists of
numbers from
the starter.
Exercise 1
4
[JMC 2015 Q19] One of the following cubes is the smallest cube that can be written as the
sum of three positive cubes. Which is it?
A 27
B 64
C 15
D 216
E 512
Solution: D
5
[JMC 2006 Q20] The sum of three different prime numbers is 40. What is the difference
between the two biggest of these numbers?
A 8
B 12
C 16
D 20
E 24
Solution: E
6
?
Important Note: If three numbers sum to an even
number, they can’t all be odd. But 2 is the only even
number, so must be one of the numbers.
[JMC 2010 Q22] Kiran writes down six different prime numbers, 𝑝, 𝑞, 𝑟, 𝑠, 𝑡, 𝑢, all less than
20, such that 𝑝 + 𝑞 = 𝑟 + 𝑠 = 𝑡 + 𝑢. What is the value of 𝑝 + 𝑞?
A 16
B 18
C 20
D 22
E 24
Solution: E
7
?
?
[TMC Regional 2009 Q9] 12345 can be expressed as the sum of two primes in exactly one
way. What is the larger of the two primes?
?
Solution: 12343. Note that odd = odd + even only. Thus one of the two primes must be 2.
Exercise 1
N
[JMO 2006 A9] The prime number 11 may be written as the
sum of three prime numbers in two different ways: 2 + 2 + 7
and 3 + 3 + 5. What is the smallest prime number which can
be written two different ways as the sum of the three prime
numbers which are all different?
Solution: 23
?
N
[JMO 2014 B6] The sum of four different prime numbers is a
prime number. The sum of some pair of the numbers is a
prime number, as is the sum of some triple of the numbers.
What is the smallest possible sum of the four prime numbers?
?
Year 7 Prime Factorisation
Prime Factorisation
To find the prime factorisation of a number is to express it as a product of
prime numbers.
?5
30 = 2 × 3 ×
12 = 22 × 3?
Bro Tip: While 12 = 2 × 2 × 3 is also correct, we can use ‘index
notation’ to group prime factors together that are the same.
= 23 × ?3 × 5
120
20
4
2
6
5
2
2
3
We can use a ‘tree’ to help us
with the working.
For each number, find two
numbers the multiply to give
it.
If you get to a prime, we
can’t branch out further, so
we have a ‘leaf’. It’s helpful to
circle the leaves.
Another quick example
2250 =
2
?
2×3
×5
2250
225
9
3
5
3
10
Possible Tree
5
5
2
?
45
3
Check Your Understanding
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 = 2 × 33? × 52
1350
1350
10
2
5
135
5
27
5
3
9
3
270
3
90 Trees
Some Possible
?3 30
3
6
5
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.
Prime Factorising a number already in index 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?
103 = 2 × 5? 3
= 2 × 5 × 2 ×? 5 × 2 × 5
= 23 × 53 ?
Quickfire Questions:
5
𝟓
?𝟓
6 =𝟐 ×𝟑
21100 = 𝟑𝟏𝟎𝟎 ×? 𝟕𝟏𝟎𝟎
1515 = 𝟑𝟏𝟓 × 𝟓?𝟏𝟓
704 = 𝟐𝟒 × 𝟓𝟒?× 𝟕𝟒
55 × 52 = 𝟓 𝟕 ?
720 × 720 = 𝟕𝟒𝟎?
2 3 × 24
= 2 × 2 × 2 × ?2 × 2 × 2 × 2
= 27 ?
Bro Note: This is an example of a ‘law
of indices’, which you will learn more
about in Year 8.
N:
Working: 910 = 310 × 310 = 320
910 = 𝟑𝟐𝟎 ?
810 = 𝟐𝟑𝟎 ?
10050 = 𝟐𝟏𝟎𝟎 ×? 𝟓𝟏𝟎𝟎
1212 = 𝟐𝟐𝟒 × ?𝟑𝟏𝟐
185 = 𝟐𝟓 × 𝟑?𝟏𝟎
12 × 2100 = 𝟑 × 𝟐𝟏𝟎𝟐
?
Exercise 2
1
By drawing a tree of otherwise, find prime
factorisations (in index form) for the following
numbers.
28 = 𝟐𝟐 × 𝟕
75 = 𝟑 × 𝟓𝟐
1000 = 𝟐𝟑 × 𝟓𝟑
378 = 𝟐 × 𝟑𝟑 × 𝟕
396 = 𝟐𝟐 × 𝟑𝟐 × 𝟏𝟏
1755 = 𝟑𝟐 × 𝟓𝟐 × 𝟏𝟑
432 = 𝟐𝟒 × 𝟑𝟑
?
?
?
?
?
?
?
2
Is 137776 odd or even?
An odd number to any power is always odd.
3
Put in prime factorised form:
35 × 36 = 𝟑𝟏𝟏
22 × 33 × 24 × 35 = 𝟐 𝟔 × 𝟑 𝟖
147 = 𝟐𝟕 × 𝟕𝟕
3658 = 𝟓𝟖 × 𝟕𝟑𝟖
5555 = 𝟓𝟓𝟓 × 𝟏𝟏𝟓𝟓
6 × 320 = 𝟐 × 𝟑𝟐𝟏
4
?
?
N1
?
?
?
?
?
?
?
Prime factorise the following:
10 000 = 𝟐𝟒 × 𝟓𝟒
257 = 𝟓𝟏𝟒
1830 = 𝟐𝟑𝟎 × 𝟑𝟔𝟎
46 × 86 = 𝟐𝟑𝟎
1 000 000100 = 𝟐𝟔𝟎𝟎 × 𝟓𝟔𝟎𝟎
?
?
?
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
𝟐 × 𝟑 or 𝟐 × 𝟑 × 𝟏 or 𝟐 × 𝟑 × 𝟏 × 𝟏.
But FLA states there is a unique
factorisation for each integer. Thus 1
is not prime.
?
N2
[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
?
Year 7 Using Prime Factorisations
Using prime factorisations
TRUE OF FALSE?
(use front of planner for true, back for false)
2 × 3 a factor of 2 × 3 × 5?
Q1
True
3
2
5
False
Remarks: You can think of 2 × 3 × 5 as
numbers in a Venn Diagram.
If we make any selection from the numbers
in this collection, we can form a factor.
Using prime factorisations
TRUE OF FALSE?
3 × 7 × 7 a multiple of 7 × 7?
Q2
True
Remarks: If 3 × 7 × 7 is a multiple of 7 × 7, then
7 × 7 must be a factor of 3 × 7 × 7.
7
7
False
3
Using prime factorisations
TRUE OF FALSE?
23 a factor of 25 ?
Q3
True
2
2
2
2
False
Remarks: Yes, if we have 2 × 2 × 2 × 2 × 2, we
can select three of these 2s to form a factor.
2
More generally, 2𝑚 will be a factor of 2𝑛 provided
that 𝑚 ≤ 𝑛.
Using prime factorisations
TRUE OF FALSE?
22 × 3 a factor of 2 × 33 ?
Q4
True
3
3
2
3
False
Remarks: No. We have one 2 and three 3s
available. 22 × 3 can’t be a factor because we
don’t have two 2s available.
Using prime factorisations
TRUE OF FALSE?
Is 15 a factor of 23 × 32 × 5?
Q5
True
False
3
2
2
3
2
5
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.
Using prime factorisations
TRUE OF FALSE?
25 × 32 × 5 a multiple of 22 × 33 ?
Q6
True
2
2
2
3
5
2
3
2
False
Remarks: We could equivalently ask whether
22 × 33 is a factor of 25 × 32 × 5.
The answer is no because there are only two 3s
available in 25 × 32 × 5, so a factor couldn’t use
three 3s.
Listing factors
TASK: List all the factors of 22 × 32
(leaving each factor in prime factorised form)
22
× ?32
22 × ?3
22 ?
2 × 3?2
2 × 3?
2 ?
32 ?
?
3
?
1
Help: Remember we have a factor of a
number if we make some selection from
all the prime factors.
2
2
3
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!)
Further Practice
1
List all the factors of 33 × 5
(leaving each factor in prime
factorised form)
33 × 5
33
32 × 5
32
?
3×5
3
5
1
2
List all the factors of 26
(leaving each factor in prime
factorised form)
26
25
24
23?
22
2
1
Number of factors
How many factors does 𝟐𝟐 × 𝟑𝟐 have?
22 × 32
22 × 3
22
2 × 32
2×3
2
32
3
1
We can see that 22 × 32 has 9 factors.
But could we have obtained this number
without having to list them all out?
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.
𝟑 × 𝟑 = 𝟗 which is what we expected!
Number of factors
How many factors does 35 × 56 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 = 𝟔 × 𝟕 = 𝟒𝟐
! 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?
22 × 35
34 × 5
2 × 76
510
1113
32 × 5 × 74
5 × 7 × 11
→
→
→
→
→
→
→
𝟑 × 𝟔 = 𝟏𝟖
103 = 𝟐𝟑 × 𝟓𝟑 → 𝟒 ×
? 𝟒 = 𝟏𝟔 factors
? factors
𝟓 × 𝟐 = 𝟏𝟎
123 = 𝟐𝟔 × 𝟑𝟑 → 𝟕 ×?𝟒 = 𝟐𝟖 factors
? factors
𝟐 × 𝟕 = 𝟏𝟒
? factors
𝟏𝟏 factors?
𝟏𝟒 factors?
𝟑×𝟐×𝟓=
? 𝟑𝟎 factors
𝟐×𝟐×𝟐=
? 𝟖 factors
Test Your Understanding
1
How many factors does 53 × 74 have?
4 × 5 = 20 factors ?
2
By first finding the prime factorisation, find the
number of factors of 120.
120 = 23 × 3 × 5
?
∴ 4 × 2 × 2 = 16 factors
3
Is 42 a factor of 22 × 5 × 73 ?
No, as 42 = 2 × 3 × 7, but 22 × 5?× 73 does not have a 3.
4
How many factors does 10050 have?
10050 = 22 × 5?2 50 = 2100 × 5100
∴ 101 × 101 = 10201 factors.
Exercise 3
1
2
3
Which of the following are factors of
53 × 72 × 11? Answer true of false.
a) 52
True
?
4
b) 5
False ?
c) 5 × 72
True
?
2
d) 7 × 13 False
?
e) 7 × 112 False
?
List all of the twelve factors of 73 × 52 ,
leaving your factors in prime factorised
form.
𝟕𝟑 × 𝟓𝟐 ,
𝟕𝟑 × 𝟓,
𝟕𝟑
𝟕𝟐 × 𝟓𝟐 ,
𝟕𝟐 × 𝟓,
𝟕𝟐
?
𝟕 × 𝟓𝟐 ,
𝟕 × 𝟓,
𝟕
𝟓𝟐 ,
𝟓,
𝟏
4
Without listing out the factors, work out
how many factors each of the following
numbers have:
a) 32 × 54
15 factors
?
2
b) 5 × 7
6 factors
?
c) 2 × 3 × 5 × 7 16 factors
?
d) 210
11 factors
?
e) 210 × 310
121 factors
?
5
Answer true or false.
6
4
7
?
a) 3 is a factor of 3
True
b) 34 × 5 is a multiple of 32 × 52 False
?
3
2
c) 35 is a factor of 2 × 3 × 5
False
?
4
2
d) 24 is a factor of 2 × 3 × 5
True, as 𝟐𝟒 = 𝟐𝟑 × 𝟑; enough 2s and
? 𝟐𝟒 × 𝟑𝟐 × 𝟓.
3s are available within
Work out how many factors each of the
following numbers have:
a) 105 = 𝟐𝟓 × 𝟓𝟓 → 𝟑𝟔 factors
?
b) 66 = 𝟐𝟔 × 𝟑𝟔 → 𝟒𝟗 factors
?
c) 1212 = 𝟐𝟐𝟒 × 𝟑𝟏𝟐 → 𝟑𝟐𝟓 factors
?
d) 100010 = 𝟐𝟑𝟎 × 𝟓𝟑𝟎 →?𝟗𝟔𝟏 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 𝟐𝟏 = 𝟑 × 𝟕 and 𝟑𝟓 = 𝟓 × 𝟕.
So the number must be some multiple of 𝟑 ×
𝟓 × 𝟕 = 𝟏𝟎𝟓. But this does have 8 factors, as
𝟐 × 𝟐 × 𝟐 = 𝟖.
?
Exercise 3
N
[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
𝟑 , or 𝒑 × 𝒒,
number has to be of the
form
𝒑
?
where 𝒑 and 𝒒 are prime numbers.
Year 7 HCF/LCM
Starter
List out the factors of 8 and 12.
What factors do they have in common?
Factors of 8:
Factors of 12:
Factors in common:
1, 2, 4, 8?
1, 2, 3, 4,
? 6, 12
1, 2, 4 ?
List out the some multiples of 8 and 12.
What multiples do they have in common?
Multiples of 8:
Multiples of 12:
Multiples in common:
8, 16, 24, 32,? 40, 48, …
? 60, 72, …
12, 24, 36, 48,
24, 48, 72, …?
Lowest Common Multiple/Highest Common Factor
Multiples of 8:
Multiples of 12:
8, 16, 24, 32, …
12, 24, 36, …
Lowest Common Multiple of 8 and 12:
24 ?
For small numbers, we can list out
multiples of the larger number until we see
a multiple of the smaller number.
Factors of 8:
Factors of 12:
1, 2, 4, 8
1, 2, 3, 4, 6, 12
Highest Common Factor of 8 and 12:
4?
For small numbers, we can list out factors
of each number and choose the greatest
number which is common.
Check Your Understanding
1
?
𝐿𝐶𝑀 60,72 = 𝟑𝟔𝟎
𝐻𝐶𝐹 60,72 = 𝟏𝟐?
Bro Shortcut: Any multiple
of 60 ends with a 0.
Therefore the multiple of 72
must be x5, x10, …
Bro Shortcut: Any number which goes
into 60 and 72 must also go into their
difference! (i.e. 12)
2
3
𝐿𝐶𝑀 12,21 = 𝟖𝟒?
𝐻𝐶𝐹 12,21 = 𝟑 ?
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.
𝑳𝑪𝑴 𝟖,?
𝟔 = 𝟐𝟒
January 25th
But what about bigger numbers?
792, 378
Sometimes it’s not practical to use this method.
Can we use the prime factorisation somehow?
3
2
2
?
3
792 =
× × 11
3?
378 = 2 × 3 × 7
But what about bigger numbers?
792 = 23 × 32 × 11
378 = 2 × 33 × 7
The ‘what wins, what loses’ method
792 = 23 × 32
× 11
378 = 2 × 33 × 7
2
𝐻𝐶𝐹 = 2? × 3?
Step 1: Align numbers so that each
prime factor has its own column.
In terms of the 2s, what do both 23 and 22 have
in common as a factor?
Step 2: To calculate HCF, see ‘what
loses’ for each of the prime factors
(i.e. lowest power, where ‘nothing’
always loses against ‘something’).
𝐿𝐶𝑀 = 2?3 × 3?3 × 7 ×? 11
= 16632
?
This time, what do both 23 and 2 go into?
Step 2: To calculate LCM, see ‘what wins’.
More Examples
2016 = 25 × 32 × 7
72 = 23 × 32
588 = 22 × 3 × 72
1232 = 24 × 7 × 11
𝑳𝑪𝑴 𝟐𝟎𝟏𝟔, 𝟕𝟐
= 𝟐𝟓 × 𝟑𝟐 × 𝟕 = 𝟐𝟎𝟏𝟔
𝑯𝑪𝑭 𝟐𝟎𝟏𝟔, 𝟕𝟐
= 𝟐𝟑 × 𝟑 = 𝟐𝟒
Line numbers up:
𝟓𝟖𝟖 = 𝟐𝟐 × 𝟑 × 𝟕𝟐
𝟏𝟐𝟑𝟐 = 𝟐𝟒
× 𝟕 × 𝟏𝟏
?
(note that if there’s a ‘draw’,
both win and both lose)
?
𝑳𝑪𝑴 𝟓𝟖𝟖, 𝟏𝟐𝟑𝟐
= 𝟐𝟒 × 𝟑 × 𝟕𝟐 × 𝟏𝟏 = 𝟐𝟓𝟖𝟕𝟐
𝑯𝑪𝑭 𝟓𝟖𝟖, 𝟏𝟐𝟑𝟐
= 𝟐𝟐 × 𝟕 = 𝟐𝟖
Check Your Understanding
?𝟐
1936 = 𝟐𝟒 × 𝟏𝟏
792 = 𝟐𝟑 × 𝟑𝟐? × 𝟏𝟏
? 𝟐 = 𝟏𝟕𝟒𝟐𝟒
𝐿𝐶𝑀 = 𝟐𝟒 × 𝟑𝟐 × 𝟏𝟏
𝐻𝐶𝐹 = 𝟐𝟑 × 𝟏𝟏? = 𝟖𝟖
If you finish…
3675 = 𝟑 × 𝟓𝟐 ×? 𝟕𝟐
875 = 𝟓𝟑 ×?𝟕
𝐿𝐶𝑀 = 𝟑 × 𝟓𝟑 × 𝟕?𝟐 = 𝟏𝟖𝟑𝟕𝟓
𝐻𝐶𝐹 = 𝟓𝟐 × 𝟕 ?= 𝟏𝟕𝟓
Exercise 4
1
Calculator permitted!
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
5
?
?
?
?
2
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
?
6
?
3
Find the LCM and HCF of the following pairs, by prime
factorising the numbers first.
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
?
?
?
?
?
?
4
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. What’s the
largest team size?
HCF(48,90) = 6
?
[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 days’ time is the party?
60 days
[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 101
B 105
C 153
D 195
Solution: B (LCM of 3, 5, 7)
?
N
𝜙 𝑛 finds the number of integers between 1 and 𝑛 that
share no factors with 𝑛 other than 1 (i.e. the HCF is 1).
For example 𝜙 6 = 2 because for two numbers up to 6,
1 and 5, HCF(6,1) = 1 and HCF(6, 5) = 1.
a) What is 𝜙(10)? 4
b) What is 𝜙(7)?
6
c) What in general is 𝜙(𝑝) for a prime number 𝑝?
𝒑−𝟏
d) Given that 𝜙 𝑚𝑛 = 𝜙 𝑚 𝜙 𝑛 (provided that
𝐻𝐶𝐹 𝑚, 𝑛 = 1), find 𝜙 1155
= 𝝓 𝟑 × 𝟓 × 𝟕 × 𝟏𝟏 = 𝝓 𝟑 𝝓 𝟓 𝝓 𝟕 𝝓 𝟏𝟏
= 𝟐 × 𝟒 × 𝟔 × 𝟏𝟎 = 𝟒𝟖𝟎
?
?
?
?
Bro Note: 𝜙(𝑛)’s posh name is ‘Euler’s Totient Function’.
Year 7 Divisibility
Divisibility Rules
How can we tell if a number is divisible by: !
2
3
4
5
6
?
Digits add up to multiple of 3. e.g: ?
1692: 1+6+9+2 = 18 
Last two digits are divisible by 4. e.g.
? 143328
Last digit is 0 or 5.
?
Number is divisible by 2 and 3 (so use
? tests for 2 and 3).
Last digit is even.
7
Double the last digit and subtract it from the remaining number, and see if the
result is divisible by 7.
e.g: 2464 -> 246 – 8 = 238 -> 23 – 16 = 7.
8
Last three digits divisible by 8.
9
Digits add up to multiple of 9.
10
Last digit 0.
11
When you sum odd-positioned digits and subtract even-positioned
digits, the result is divisible by 11. ?
?
?
?
?
e.g. 47949: (4 + 9 + 9) – (7 + 4) = 22 – 11 = 11, which is divisible by 11.
12
Number divisible by 3 and by 4.
?
Quickfire Divisibility
4
6 7
? ?
?
726
168
? ? ?
9196 ? ? ?
252
? ? ?
?
? ?
1001
?
? ?
91
216
? ? ?
87912 ? ? ?
9
11
? ?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
Quickfire Mental Primes
Apart from the obvious instant checks (divisibility by 2, 5), we
usually only have to mentally check 3, 7 and 11 to have a good
‘guess’ that a number is prime.
3
91
101
234567 
131
781
751
221
7
?
11 Is it prime?
No ?
Yes ?
?
?
?
? 
?
?
?
?
?
Yes ?
No! (13? × 17)
No
Yes
No
Advanced: 13 trick is
“Quadruple last digit
and add to remaining
number. Is result
divisible by 13?
N For 221, what is the largest prime we would have had to test divisibility until we’d be certain it was prime?
Up to 𝟐𝟐𝟏 = 𝟏𝟒. 𝟖𝟕 because all composite numbers have a factor (other than 1) up to the square root.
?
Test Your Understanding
Easier One:
[JMO 1997 A5] Precisely, one of the numbers 234, 2345, 23456, 234567, 2345678,
23456789 is a prime number. Which one must it be?
Solution: 23456789?
Harder One:
[JMC 2012 Q23] Peter wrote a list of all the numbers that could be produced by
changing one digit of the number 200. How many of the numbers on Peter’s list are
prime?
A 0
B 1
C 2
D 3
E 4
Solution: A
?
Breaking Down Divisibility Rules
Are these statements true or false?
If we want to show that a number
is divisible by 15, we can show it
is divisible by 3 and 5.
False 
True 
If we want to show that a number
is divisible by 24, we can show it
is divisible by 6 and 4.
False 
True 
The problem is that 12 is divisible by 6 and 4, but it is not divisible
by 24!
We need to pick two numbers which are coprime, i.e. do not
share any factors.
How can we therefore test if a number is divisible by 24?
! Two numbers 𝑎 and 𝑏 are coprime if 𝑯𝑪𝑭 𝑎, 𝑏 = 1
Quickfire
What divisibility rules would we use if we wanted to test divisibility by:
18
45
36
40
2 and 9 rules
5 and 9
4 and 9
5 and 8
?
?
?
?
An easy Year 10 Maths Olympiad problem:
Find the smallest positive integer
which consists only of 0s and 1s,
and which is divisible by 12.
• Since in must be divisible by 4, the only
possibility for the last two digits is 00.
• It must have at least?three 1s to be divisible
by 3 (as we can’t have zero 0s).
• Therefore 11100 is the answer.
Exercise 5
Problem sheet of Junior and Intermediate Olympiad problems.
Work in pairs/groups if you wish.
Answers on next slides.
(See printout)
Question 1
[J31] Every digit of a given positive integer is either a 3 or a 4 with each occurring at least
once. The integer is divisible by both 3 and 4. What is the smallest such integer?
?
Question 2
[J50] The eight-digit number “ppppqqqq”, where p and q are digits, is a multiple of 45.
What are the possible values of p?
?
Question 3
[M07] (a) A positive integer N is written using only the digits 2 and 3, with each appearing
at least once. If N is divisible by 2 and by 3, what is the smallest possible integer N?
(b) A positive integer M is written using only the digits 8 and 9, with each appearing at
least once. If M is divisible by 8 and by 9, what is the smallest possible integer M?
?
?
Question 4
[M55] A palindromic number is one which reads the same when its digits are reversed, for
example 23832. What is the largest six-digit palindromic number which is exactly divisible
by 15?
?
Question 5
[J16] Find a rule which predicts exactly when five consecutive integers have sum divisible
by 15.
?
Question 6
[M96] Find the possible values of the digits p and q, given that the five-digit number
‘p543q’ is a multiple of 36.
?
Question 7
[M127] The five-digit number ‘𝑎679𝑏’, where 𝑎 and 𝑏 are digits, is divisible by 36. Find all
possible such five-digit numbers.
?
Question 8
[M31] Find the smallest positive multiple of 35 whose digits are all the same as each
other.
?
Question 9
[Based on NRich] If the digits 5, 6, 7 and 8 are inserted at random in 3_1_4_0_92 (one in
each space), what is the probability that the number created will be a multiple of 396 if:
a) Each of 5, 6, 7, 8 is used exactly once in each of the four gaps.
b) Each of 5, 6, 7, 8 can be used multiple times.
[JAF solution] If number is divisible by 396, it is divisible by 4 × 9 × 11.
If we used one each of 5, 6, 7, 8 to fill the gaps, all three divisibility rules, for 4, 9 and
11, would be satisfied regardless of order, since the last two digits are fixed and all
inserted digits are in even positions. Thus the probability is 1.
b) If we used 5, 6, 7, 8, there are 4! = 24 possible orderings.
However, we could also use 5, 5, 8, 8 to fill the gaps, as this will not affect the digit sums
involved in either the divisibility by 9 or 11 rules (since all digits are inserted in even
?
positions), and the last two digits are fixed for the purposes of the 4 rule. There are 6
possible orderings of these digits.
6, 6, 7, 7 is also possible, which by the same reasoning, gives 6 possible orderings.
5, 7, 7, 7 is also possible, which gives 4 orderings.
6, 6, 6, 8 is also possible, which gives 4 orderings.
There are therefore 4 + 4 + 6 + 6 + 24 = 44 possible numbers divisible by 396.
However, there are 44 total possible ways of using 5, 6, 7 or 8 for each of the four gaps.
44
11
The probability is therefore 44 = 64
N BONUS QUESTION
Show that:
𝑛3 − 𝑛 is divisible by 6 for all integers 𝑛.
𝑛4 + 2𝑛3 − 𝑛2 − 2𝑛 is divisible by 24 for all integers 𝑛.
a) 𝑛3 − 𝑛 = 𝑛 𝑛 − 1 𝑛 + 1
This is the product of three consecutive numbers. One of the three
numbers must be divisible by 3, so the product is divisible by 3.
Similarly, at least one of the three numbers is divisible by 2, so the
product is divisible by 2. Therefore, the product is divisible by 6.
b) = 𝑛3 𝑛 + 2 − 𝑛 𝑛 + 2
= 𝑛3 − 𝑛 𝑛 + 2
?
= 𝑛−1 𝑛 𝑛+1 𝑛+2
This is the product of 4 consecutive numbers.
At least one of the four is divisible by 3.
Exactly two of the numbers will be divisible by 2.
However, one of the four numbers will be divisible by 4, giving an
extra factor of 2. Overall, this means the product is divisible by
2 × 3 × 4 = 24.
Fini
(any slides after this are supplementary)
Year 7 Divisibility Within
Equations
STARTER :: Divisibility of Expressions
Given that 𝑛 is an integer, for each expression, identify whether
the statement is always true, always false, or sometimes true.
Always true
𝟓𝒏 is divisible by 5.
Always false

?
(5 times a number is clearly
in the 5 times table)

𝟒𝒏 + 𝟏 is divisible
by 4.
𝟒 𝒏 + 𝟏 is
divisible by 4.
?
(This will always be one more
than a multiple of 4)

𝟑𝒏 is divisible by 2.
?
?
𝟔𝒏 is divisible by 2.

N 𝒏 𝒏 + 𝟏 is
divisible by 2.

N 𝟒𝒏𝟐 is a square
Sometimes true
(one of 𝑛 and 𝑛 + 1 is
even, and
𝑜𝑑𝑑 × 𝑒𝑣𝑒𝑛 = 𝑒𝑣𝑒𝑛)

(The product of two
squares is a square)
?
?
?

Factors within equations
What can you tell about the following numbers in these
equations?
? 𝒐𝒇 𝟑
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3 + 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3 = 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒆
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 4 + 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒆
? 𝒐𝒇 𝟒 = 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 4
What can you therefore say about 4𝑦 in this equation, if 𝑥
and 𝑦 are integers?
5𝑥 + 𝟒𝒚 = 150
? by 5, 𝟒𝒚 must be.
Since 𝟓𝒙 and 150 are divisible
What therefore can you say about 𝑦?
4 is not divisible by 5. ?Therefore 𝒚 must be.
Divisibility in Equations
Find positive integer solutions to:
5𝑥 + 4𝑦 = 50
Bro Side Note: The posh name for this
is a ‘linear Diophantine equation’. The
‘Diophantine’ bit means we’re only
looking for integer solutions. The
‘linear’ bit means that if we plotted the
solutions (𝑥, 𝑦) on some axes, they’d
form a straight line.
One strategy would be to try different values of 𝑥 and see if it works for 𝑦. e.g.
If 𝑥 = 1:
𝟒𝒚 would be 45. But 45 doesn’t divide by 4.
?
Can you think of a more intelligent way?
𝟓𝒙 and 50 are multiples of 5, therefore 𝟒𝒚 is a multiple of 5, and therefore 𝒚
?
is a multiple of 5.
Therefore 𝑦 can only be:
𝒚 = 𝟓, 𝟏𝟎
? have to be negative.
If 𝒚 was 15 then 𝟒𝒚 = 𝟔𝟎, then 𝒙 would
If 𝑦 = 5 then 𝒙 = 𝟔?
If 𝑦 = 10 then 𝒙 = 𝟐?
Bro Tip: It’s sometimes convenient to
write the solutions as 𝑥, 𝑦 = 5,6 , …
More examples
Find positive integer solutions to:
7𝑥 + 6𝑦 = 90
As 6𝑦 and 90 are divisible by 6, 7𝑥 needs to be divisible by 6, therefore 𝑥 is
divisible by 6.
𝑥 can be 6 or 12 (as next value of 𝑥 is 18,
? and 7𝑥 would be greater than 90).
If 𝑥 = 6, 𝑦 = 8
If 𝑥 = 12, 𝑦 = 1
Bro Note: What is good about this method is that we have not
only found all the solutions: we’ve shown there can’t be any
others. This is very important in maths!
Does the following equation have any integer solutions?
2𝑥 + 4𝑦 + 1 = 6𝑇
No. The left-hand-side of the equation is odd but the right-hand-side of the
equation is even.
?
Test Your Understanding
Easier One:
[JMC 2010 Q12] Sir Lance has a lot of tables and chairs in his house. Each rectangular table
seats eight people and each round table seats five people. What is the smallest number of
tables he will need to use to seat 35 guests and himself, without any of the seating around
these tables remaining unoccupied?
(Hint: if 𝑥 is the number of rectangular tables, and 𝑦 is number of circular tables, form an equation first that looks like
what we’ve previously seen)
A 4
B 5
𝟖𝒙 + 𝟓𝒚 = 𝟑𝟔
C 6
D 7
E 8
? 𝒚 is. When 𝒚 = 𝟒, 𝒙 = 𝟐. So answer is C
𝟓𝒚 is divisible by 4 therefore
Harder One:
[JMC 2012 Q25] The interior angles of a triangle are
5𝑥 + 3𝑦 °, 3𝑥 + 20 ° and 10𝑦 + 30 °, where 𝑥, 𝑦
are positive integers. What is the value of 𝑥 + 𝑦?
𝟓𝒙 + 𝟑𝒚 + 𝟑𝒙 + 𝟐𝟎 + 𝟏𝟎𝒚 + 𝟑𝟎 = 𝟏𝟖𝟎
𝟖𝒙 + 𝟏𝟑𝒚 + 𝟓𝟎 = 𝟏𝟖𝟎
𝟖𝒙 + 𝟏𝟑𝒚 = 𝟏𝟑𝟎 ?
𝒙 must be divisible by 13. 13 is the only possibility.
If 𝒙 = 𝟏𝟑, 𝒚 = 𝟐. Therefore 𝒙 + 𝒚 = 𝟏𝟓
Exercise 6
1 Find all positive integer solutions to the
following:
a) 2𝑥 + 3𝑦 = 10
𝒙, 𝒚 = 𝟐,?𝟐
b) 4𝑥 + 5𝑦 = 35
𝒙, 𝒚 =? (𝟓, 𝟑)
c) 7𝑥 + 5𝑦 = 77
d) 17𝑥 + 5𝑦 = 340
3
?
4
𝒙, 𝒚 = 𝟔,?𝟕 , (𝟏, 𝟏𝟒)
𝒙, 𝒚 = 𝟏𝟓, 𝟏𝟕 ,
?
𝟏𝟎, 𝟑𝟒 , (𝟓, 𝟓𝟏)
2 [JMC 2005 Q16] ‘Saturn’ chocolate bars
are packed either in boxes of 5 or boxes of
12. What is the smallest number of full
5
boxes required to pack exactly 2005
‘Saturn’ bars?
A 118 B 167 C 168
D 170 E 401
Solution: D
?
[JMO 2012 A4] A book costs £3.40 and a
magazine costs £1.60. Clara spends
exactly £23 on books and magazines.
How many magazines does she buy?
Solution: 8
[IMC 2001 Q16] The Pythagoras
Patisseries sells triangular cakes at 39p
each and square buns at 23p each. For
her party, Helen spent exactly £5.12 on
an assortment of these cakes and buns.
How many items in total did she buy?
A 15
B 16
C 17
D 18
E 19
Solution: B
?
If 𝑥 is divisible by 3, what can you say
about?
a) 𝑥 2 ? Divisible by 9
b) 4𝑥 2 ? Divisible by 36
c) 4 𝑥 + 1 ? Divisible by 4
c) 2𝑥 2 3𝑥 + 1 ? Divisible by 18
?
?
?
?
Year 7 Square and cube numbers
Square/Cube Numbers
Find the prime factorisation of the following square numbers. What do you notice?
49 = 𝟕𝟐 ?
324 = 𝟐𝟐 × 𝟑
?𝟒
18225 = 𝟑𝟔 × 𝟓?𝟐
! Square numbers have even powers in their prime factorisation.
Find the prime factorisation of the following cube numbers. What do you notice?
8 = 𝟐𝟑 ?
8000 = 𝟐𝟔 × ?
𝟓𝟑
19683 = 𝟑𝟗 ?
!Cube numbers have powers which are multiples of 3.
Square/Cube Numbers
Q
What is the smallest multiple of 504 that is square?
(Note that 504 = 23 × 32 × 7)
The power of 3 on the 2 and the power of 1 on the 7 are both odd.
?
So we need to multiply by 2 and 7, i.e. 14, to give 7056.
a
Check your understanding:
What do you need to multiply the following by to make the following square and cube?
Square
b
Cube
504 = 23 × 32 × 7
× 14
× 3 × 7?2 (147)
12 = 22 × 3
×?3
×2×3
?2 (18)
800 = 25 × 52
×?2
× 2 ×?
5 (10)
How many square numbers are factors of 210 ?
Bro Reminder: Factors of 210 will be all
powers of 2 with a power up to 10.
All even powers of 2 (𝟐𝟐 , 𝟐𝟒 , 𝟐𝟔 , 𝟐𝟖 , 𝟐𝟏𝟎 ). But don’t forget that 𝟏 (= 𝟐𝟎 ) is square!
(as the 0 is even). That’s 6 square factors.?
Numbers of factors of square numbers
List the factors of the (non-square) number 20.
How many factors are there?
1, 2, 4, 5, 10, 20 (6 factors)
?
List the factors of the (square) number 49.
How many factors are there?
1, 7, 49 (3 factors)
?
The number of factors of a square number will always be odd. Can you
explain why this is?
Normally factors come in pairs. e.g. For 20, we have 𝟏 × 𝟐𝟎, 𝟐 × 𝟏𝟎, 𝟒 × 𝟓
However, in 𝟒𝟗, the 7 in 𝟕 × 𝟕 only counts
? once, so we’ll have an odd
number of factors.
! Square numbers (and only square numbers) have an odd number of factors.
Exercise 7 (on provided sheet)
1
Use the following prime factorisations (if
provided) to find the smallest number we
need to multiply the number by to make it (a)
a square (b) a cube.
(i) 150 = 2 × 3 × 52 (a) 6 (b) 180
(ii) 24 = 23 × 3
(a) 6 (b) 9
(ii) 16200
(a) 2 (b) 45
?
?
?
2
List the cube numbers (in prime factorised
form) that are factors of 512 .
Solution: 1, 𝟓𝟑 , 𝟓𝟔 , 𝟓𝟗 , 𝟓𝟏𝟐
?
3
[IMC 1999 Q16] Three statements:
i) 310 is even (ii) 310 is odd (iii) 310 is square
Exactly which ones are true?
A (i) only
B (ii) only
C (iii) only
D (i) and (iii) only
E (ii) and (iii)
Solution: E
?
4
5
[JMO 2008 A6] How many positive square
numbers are factors of 1600?
Solution: 8
How many numbers between 1 and 16 have an
odd number of factors?
Solution: 4 (the four squares: 1, 4, 9, 16)
?
?
6 [Junior Kangaroo 2015 Q23] How many three-
digit numbers have an odd number of factors?
A 5 B 10
C 20 D 21 E 22
E (22 squares between 100 and 961)
?
N1 [Cayley 2013 Q1] What is the smallest non-zero
multiple of 2, 4, 7 and 8 which is a square?
Solution: 784
?
N2 [JMO 2012 B2] Anastasia thinks of a positive
integer, which Barry then doubles. Next, Charlie
trebles Barry's number. Finally, Damion
multiplies Charlie's number by six. Eve notices
that the sum of these four numbers is a perfect
square. What is the smallest number that
Anastasia could have thought of? (Hint: make
Anastasia’s number 𝑥)
Solution: Numbers are 𝒙, 𝟐𝒙, 𝟔𝒙, 𝟑𝟔𝒙 which
add up to 𝟒𝟓𝒙. Since 𝟒𝟓 = 𝟑𝟐 × 𝟓, then 𝒙
could be 5 to give 𝟑𝟐 × 𝟓𝟐 which is square.
Thus Anastasia’s number was 5.
?
Supplementary Slides on Divisibility
“𝟓𝒙 is divisible by 15”.
Try different values of 𝑥 to see for what values
this is true.
For example, if 𝑥 = 4, then 5𝑥 is 20, but 20 is
not divisible by 15.
What property does 𝑥 have when it works?
?
𝒙 must be divisible
by 3.
Thinking in buckets
“3𝑥 is divisible by 2”
We can think of the 3 and 𝑥 of 3𝑥 as
‘buckets’ which we can put prime factors in
if it is a factor of the number/expression.
The 3 bucket can’t have the
factor of 2, so it must have
been in the 𝑥 bucket.
3
𝑥
Therefore 𝑥 is divisible by 2.
2
Factors
Click for
Bromanimation
Thinking in buckets
“6𝑥 is divisible by 3”
The 3 may have come from the
6 bucket (6 is divisible by 3).
6
3
Factors
𝑥
But it may or may not have
come from the 𝑥 bucket
(depending on whether 𝑥 is
divisible by 3)
Because the prime factor of
3 could have just come from
the 6, we can’t guarantee
anything about the
divisibility of 𝑥.
Click for
Bropossibility 1
Click for
Bropossibility 2
Thinking in buckets
We have a factor of 2 and 3
to put in the buckets. The 2
must have come from the 𝑥
bucket, but the 3 could have
come from the ‘3 bucket’. In
such a case, we can
guarantee 𝑥 is divisible by 2.
“3𝑥 is divisible by 6”
3
2 ?3
Factors
𝑥
But the 3 might have also
been in the 𝑥 bucket. 𝑥
would be divisible by 6.
There what can we
guarantee about 𝑥?
It must be divisible by 2,
but we can’t guarantee it
? by 6
will be divisible
(because of the first case).
Click for
Bropossibility 1
Click for
Bropossibility 2
Thinking in buckets
“𝑥 2 is divisible by 3”
3
𝑥
𝑥
This one’s more
complicated. The 3 has to go
in one of the buckets. But
the two buckets represent
the same number (𝑥), so we
know both buckets must
have a 3.
We know therefore:
• 𝑥 is divisible by 3.
• 𝑥 2 is divisible by 9.
3
Factors
Click for
Bromanimation
Card Sort
Match the statements with the strongest statement they MUST result in.
(By strongest, I mean for example that “is divisible by 8” is stronger than “is divisible by 2” as it is more restrictive)
Some cards may not be used and some orange cards may match multiple green.
5𝑥 is divisible by 5.
𝑥 is divisible by 2.
10𝑥 is divisible by 5.
𝑥 is divisible by 3.
5𝑥 is divisible by 10.
10𝑥 is divisible by 15.
𝑥 2 is divisible by 5
𝑥 is divisible by 4.
𝑥 is divisible by 5.
𝑥 is divisible by 6.
2𝑥 2 is divisible by 4
𝑥 is divisible by 25.
2𝑥 3 is divisible by 6
𝑥 is divisible by 27.
3𝑥 3 4𝑥 + 1 is divisible
by 24
We don’t know
anything about 𝑥
Further Divisibility of Expressions
𝑥 is divisible by 2. What is the largest number we can say
𝑥 2 4𝑥 + 2 is divisible by?
Try a few values of 𝑥 first and see.
𝑥 𝑥
2 2
4𝑥 + 2
2
Factors
Answer: 8 ?
If we add a multiple
of 4 and a multiple
of 2, it will give a
multiple of 2 but
not of 4.
More Examples
If 𝑛 is divisible by 3, what can we say about the divisibility of…
a
𝑛(𝑛 + 2)(𝑛 + 3)
𝒏 and 𝒏 + 𝟑 are in the 3 times table, but 𝒏 + 𝟐 will be two
more than a multiple of 3. ?
So the expression is divisible by 𝟑 × 𝟑 = 𝟗.
Test Your Understanding:
b
𝑛3 6𝑛 + 2
We get a factor of 3 three times from 𝒏𝟑 . 𝟔𝒏 + 𝟐 is not
divisible by 3, but is even, giving
? us a factor of 2.
So the expression is divisible by 𝟑 × 𝟑 × 𝟑 × 𝟐 = 𝟓𝟒.
Supplementary Questions (for teacher use)
1
What are the factors of 524 that are both square and
cube? Leave your answer in factorised index form.
Solution: 1, 𝟓𝟔 , 𝟓𝟏𝟐 , 𝟓𝟏𝟖 , 𝟓𝟐𝟒
6
?
2
3
How many positive factors does 1212 have?
Solution: 325
?
A certain number has exactly eight factors including
1 and itself. Two of its factors are 33 and 15. What is
the number? Solution: 165
?
4
[SMC 2003 Q15] The number of this year, 2003, is
prime. How many square numbers are factors of
20032003 ?
Solution: 1002
?
5
[Kangaroo Pink 2010 Q18] How many integers 𝑛,
between 1 and 100 inclusive, have the property that
𝑛𝑛 is a square number?
A 99 B 55 C 50 D 10 E 5
Solution: B
?
[JMO Mentoring Jun2011 Q2] How
many positive divisors does 6! have
including 6! and 1? [6! = 6 × 5 × 4 ×
3 × 2 × 1.] Solution: 30
?
7
[JMO 1996 B1] How many positive
whole numbers up to and including 400
can be written in exactly one way as the
product of two even numbers?
Solution: 25 (each product must be of
the form 𝟐 × 𝟐𝒑 where 𝒑 is prime)
?
BONUS LESSON! Use #3: Trailing zeroes
Prime Factorisation Number
Trailing Zeroes
23 × 3 × 54
15 000
3
25 × 52 × 7
5600
2
48 400 000
5
27 × 55 × 112
Can you think of a rule that tells us the number of trailing zeroes from the prime
factorisation? Why do you think it works?
It is the lowest power of the 2 and 5. This is because each 2-5 pair forms a factor
of 10, which puts a 0 on the end of the number.
The lower power tells us how
?
many pairs we can make.
Examples
Q
Q
Q
How many zeroes are on the end of 2100 × 340 × 580 ?
80 trailing zeroes.
?
8! is said “8 factorial” and means 8! = 8 × 7 × ⋯ × 2 × 1
How many trailing zeroes does 8! have?
𝟖! = 𝟐𝟑 × 𝟕 × 𝟐 × 𝟑 × 𝟓 × 𝟐𝟐 × 𝟑 × 𝟐
= 𝟐𝟕 × 𝟑𝟐 × 𝟓 × 𝟕
Thus there is only 1 trailing zero.
? Note that 𝒏! seems to
give us a lot more twos than fives, so we only need to
count the number of fives to get that trailing zeroes.
Without using your calculator, what is the last non-zero digit
of 27 × 32 × 55 ?
If we get rid of all the trailing zeroes, the last digit will be
the one we want. We get rid ?
of the five 2-5 pairs to leave
𝟐𝟐 × 𝟑𝟐 . This is 36, so the last non-zero digit is 6.
Test Your Understanding
a
b
N
How many zeroes are on the end of 240 × 590 × 1120 ?
40 trailing zeroes.
?
What is the last non-zero digit of 280 × 579 × 7?
Getting rid of trailing zeroes leaves 𝟐 × 𝟕 = 𝟏𝟒, so last
?
non-zero digit is 4.
How many zeroes are at the end of 25! (remember that we
only care about the number of 5s in the prime factorisation)
Each multiple of 5 in 25 x 24 x … gives us a prime factor of
5, and there’s 5 of them. However
25 gives us an extra 5.
?
So we have six 5s, and lots of 2s, giving six trailing zeroes.
Exercise 8 (on provided sheet)
1
2
How many zeroes are at the end of 230 × 531 ?
Solution: 31
How many zeroes are at the end of 240 ×
330 × 520 ?
Solution: 20
?
?
3
4
What is the last non-zero digit of:
210 × 32 × 511
Solution: 5
28 × 57 × 72
Solution: 8
10
8
2 × 5 × 13
Solution: 2
?
?
?
[JMO 2010 A3] Tom correctly works out 2010
and writes down his answer in full. How many
digits does he write down in his full answer?
Solution: 11
[IMC 2007 Q7] If the numbers 1, 2, 3, 4, 5, 6, 7,
8, 9, 10 are all multiplied together, how many
zeros are at the end of the answer?
Solution: 2
[IMC 2000 Q18] The number 34 × 45 × 56 is
written out in full. How many zeroes are there
at the end of the number?
Solution: 6
?
5
?
6
?
7 [Kangaroo Pink 2012 Q16] What is the last
non-zero digit when 257 × 34 × 553 is
evaluated?
Solution: 6
8 [Kangaroo Grey 2004 Q25] The number 𝑁
is the product of the first 100 positive
whole numbers. If all the digits of 𝑁 were
written out, what digit would be next to
all the zeros at the end?
A 2
B 4
C 6
D 8
E 9
Solution: 4
Sam correctly calculates
9 [SMC 2001 Q15]
the value of 58 × 85 . How many digits
does her answer contain?
Solution: 11
10 [Senior Kangaroo 2012 Q1] How many
zeroes are there at the end of the number
which is the product of the first 2012
prime numbers?
Solution: 1
11 Find two integers, neither of which has a
zero digit, whose product is 1 000 000.
𝟑𝟐 × 𝟑𝟏𝟐𝟓
?
?
?
?
?