1. Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com
Year 7 Number Theory
Dr J Frost
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: 22nd July 2018

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Questions organised by topic, difficulty and past paper.
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3. 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
Extension 1: Divisibility of Terms/Within Equations
Extension 2: More on Squares and Cubes
Even More Stuff!
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4. 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, 121, 144, 169, 196, 225, 256
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:
(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)
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66 ? ? ? ?
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.

5. ! ! 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 is a ‘perfect square’, but is not necessarily a square number, e.g. if 𝑥=0.5 (as =2.25) ?

6. 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.
How many ways are there of expressing 100 as the sum of two primes?
Solution: 6 (3 + 97, , , , , )
Bro Tip: It often helps to write 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 𝟐 𝟓 𝟐 = 𝟕 𝟐 +𝟐 𝟒 𝟐 =𝟐 𝟎 𝟐 +𝟏 𝟓 𝟐 ) ?

7. Exercise 1 ? ? ? (Problems on provided sheet)
[JMC 2015 Q11] What is the smallest prime number that is the sum of three different prime numbers? A B C D E 23 Solution: D
[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)
[JMO 2009 A3] The positive whole numbers 𝑎, 𝑏 and 𝑐 are all different and 𝑎 2 + 𝑏 2 + 𝑐 2 =121. What is the value of 𝑎+𝑏+𝑐?
Solution: 17
Bro Tip: Again, use your lists of numbers from the starter. 1 ? 2 ? 3 ?

8. Exercise 1
[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 B C D E Solution: D
[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 C D E Solution: E
[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 B C D E Solution: E
[TMC Regional 2009 Q9] can be expressed as the sum of two primes in exactly one way. What is the larger of the two primes? Solution: Note that odd = odd + even only. Thus one of the two primes must be 2. 4 ? 5
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. ? 6 ? 7 ?

9. Exercise 1
[JMO 2006 A9] The prime number 11 may be written as the sum of three prime numbers in two different ways: and 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
[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? N ? N ?

10. Year 7 Prime Factorisation

11. Prime Factorisation 30=2×3×5 12= 2 2 ×3 120 20 6 4 ? ? ? = 2 3 ×3×5 5
To find the prime factorisation of a number is to express it as a product of prime numbers. ?
30=2×3×5 12= 2 2 ×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. ?
120
= 2 3 ×3×5
We can use a ‘tree’ to help us with the working.
For each number, find two numbers the multiply to give it. 20 6 4 5 2 3
If you get to a prime, we can’t branch out further, so we have a ‘leaf’. It’s helpful to circle the leaves. 2 2

12. 2250=2× 3 2 × 5 3 Another quick example 2250 225 10 45 9 ?
Possible Tree ?
2250
225 10 45 5 2 5 9 5 3 3

13. 1350=2× 3 3 × 5 2 Check 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=2× 3 3 × 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.

14. 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?
10 3 = 2× =2×5×2×5×2× = 2 3 × 5 3 ?
2 3 × 2 4 = 2×2×2 × 2×2×2×2 = 2 7 ? ? ? ?
Bro Note: This is an example of a ‘law of indices’, which you will learn more about in Year 8.
Quickfire Questions: N:
9 10 = 𝟑 𝟐𝟎
8 10 = 𝟐 𝟑𝟎
= 𝟐 𝟏𝟎𝟎 × 𝟓 𝟏𝟎𝟎
12 12 = 𝟐 𝟐𝟒 × 𝟑 𝟏𝟐
1 8 5 = 𝟐 𝟓 ×𝟑 𝟏𝟎
12× =𝟑× 𝟐 𝟏𝟎𝟐
Working: = 3 10 × 3 10 = 3 20 ?
6 5 = 𝟐 𝟓 × 𝟑 𝟓
= 𝟑 𝟏𝟎𝟎 × 𝟕 𝟏𝟎𝟎 = 𝟑 𝟏𝟓 × 𝟓 𝟏𝟓
70 4 = 𝟐 𝟒 × 𝟓 𝟒 × 𝟕 𝟒
5 5 × 5 2 = 𝟓 𝟕
7 20 × 7 20 = 𝟕 𝟒𝟎 ? ? ? ? ? ? ? ? ? ? ?

15. Exercise 2
By drawing a tree of otherwise, find prime factorisations (in index form) for the following numbers.
28= 𝟐 𝟐 ×𝟕 75=𝟑× 𝟓 𝟐 1000= 𝟐 𝟑 × 𝟓 𝟑 378=𝟐× 𝟑 𝟑 ×𝟕 396= 𝟐 𝟐 × 𝟑 𝟐 ×𝟏𝟏 1755= 𝟑 𝟐 × 𝟓 𝟐 ×𝟏𝟑 432= 𝟐 𝟒 × 𝟑 𝟑
Is odd or even?
An odd number to any power is always odd.
Put in prime factorised form: 3 5 × 3 6 = 𝟑 𝟏𝟏 × 3 3 × 2 4 × 3 5 = 𝟐 𝟔 × 𝟑 𝟖 = 𝟐 𝟕 × 𝟕 𝟕 = 𝟓 𝟖 ×𝟕 𝟑 𝟖 = 𝟓 𝟓𝟓 × 𝟏𝟏 𝟓𝟓 6× 3 20 =𝟐× 𝟑 𝟐𝟏 1 4
Prime factorise the following:
10 000= 𝟐 𝟒 × 𝟓 𝟒 = 𝟓 𝟏𝟒 = 𝟐 𝟑𝟎 × 𝟑 𝟔𝟎 × 8 6 = 𝟐 𝟑𝟎 = 𝟐 𝟔𝟎𝟎 × 𝟓 𝟔𝟎𝟎
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.
[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 ? ? ? ? ? ? ? ? ? ? N1 ? ? 2 ? ? 3 ? ? ? N2 ? ? ? ?

16. Year 7 Using Prime Factorisations

17. 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
2×3 a factor of 2×3×5?
True
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. 3 2 5

18. True False TRUE OF FALSE? Using prime factorisationsQ2
3×7×7 a multiple of 7×7?
True
False
Remarks: If 3×7×7 is a multiple of 7×7, then 7×7 must be a factor of 3×7×7. 7 3 7

19. True False TRUE OF FALSE? Using prime factorisationsQ3
2 3 a factor of 2 5 ?
True
False 2 2
Remarks: Yes, if we have 2×2×2×2×2, we can select three of these 2s to form a factor.
More generally, 2 𝑚 will be a factor of 2 𝑛 provided that 𝑚≤𝑛. 2 2 2

20. True False TRUE OF FALSE? Using prime factorisationsQ4
2 2 ×3 a factor of 2× 3 3 ?
True
False 3
Remarks: No. We have one 2 and three 3s available ×3 can’t be a factor because we don’t have two 2s available. 3 2 3

21. True False TRUE OF FALSE? Using prime factorisationsQ5
Is 15 a factor of 2 3 × 3 2 ×5?
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

22. True False TRUE OF FALSE? Using prime factorisationsQ6
2 5 × 3 2 ×5 a multiple of 2 2 × 3 3 ?
True
False 2 2
Remarks: We could equivalently ask whether 2 2 × 3 3 is a factor of 2 5 × 3 2 ×5.
The answer is no because there are only two 3s available in 2 5 × 3 2 ×5, so a factor couldn’t use three 3s. 2 3 5 2 3 2

23. Listing factors TASK: List all the factors of 2 2 × 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 × × × × ? ? ? 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!) ?

24. Further Practice 1
List all the factors of 3 3 ×5 (leaving each factor in prime factorised form) 2
List all the factors of (leaving each factor in prime factorised form) ?
3 3 × × × ?

25. How many factors does 𝟐 𝟐 × 𝟑 𝟐 have?
Number of factors
How many factors does 𝟐 𝟐 × 𝟑 𝟐 have?
We can see that 2 2 × 3 2 has 9 factors. But could we have obtained this number without having to list them all out?
2 2 × × × ×
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! ?

26. How many factors does 3 5 × 5 6 have?
Number of factors
How many factors does 3 5 × 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 =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?
2 2 × → 𝟑×𝟔=𝟏𝟖 factors 3 4 ×5 → 𝟓×𝟐=𝟏𝟎 factors
2× → 𝟐×𝟕=𝟏𝟒 factors
→ 𝟏𝟏 factors
→ 𝟏𝟒 factors
3 2 ×5× 7 4 → 𝟑×𝟐×𝟓=𝟑𝟎 factors
5×7×11 → 𝟐×𝟐×𝟐=𝟖 factors ?
10 3 = 𝟐 𝟑 × 𝟓 𝟑 → 𝟒×𝟒=𝟏𝟔 factors = 𝟐 𝟔 × 𝟑 𝟑 →𝟕×𝟒=𝟐𝟖 factors ? ? ? ? ? ? ? ?

27. Test Your Understanding1
How many factors does 5 3 × 7 4 have? ?
4×5=20 factors 2
By first finding the prime factorisation, find the number of factors of 120. ?
120= 2 3 ×3×5
∴4×2×2=16 factors 3
Is 42 a factor of 2 2 ×5× 7 3 ? ?
No, as 42=2×3×7, but 2 2 ×5× 7 3 does not have a 3. 4
How many factors does have? ?
= × = × 5 100
∴101×101=10201 factors.

28. Exercise 3
Without listing out the factors, work out how many factors each of the following numbers have:
3 2 × factors
5 2 ×7 6 factors
2×3×5× factors
factors
2 10 × factors
Work out how many factors each of the following numbers have:
= 𝟐 𝟓 × 𝟓 𝟓 →𝟑𝟔 factors
= 𝟐 𝟔 × 𝟑 𝟔 →𝟒𝟗 factors
= 𝟐 𝟐𝟒 × 𝟑 𝟏𝟐 →𝟑𝟐𝟓 factors
= 𝟐 𝟑𝟎 × 𝟓 𝟑𝟎 →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.
Which of the following are factors of × 7 2 ×11? Answer true of false.
True
False
5× True
7 2 × False
7× False
List all of the twelve factors of 7 3 × 5 2 , leaving your factors in prime factorised form.
𝟕 𝟑 × 𝟓 𝟐 , 𝟕 𝟑 ×𝟓, 𝟕 𝟑 𝟕 𝟐 × 𝟓 𝟐 , 𝟕 𝟐 ×𝟓, 𝟕 𝟐 𝟕× 𝟓 𝟐 , 𝟕×𝟓, 𝟕 𝟓 𝟐 , 𝟓, 𝟏
Answer true or false.
3 4 is a factor of True
3 4 ×5 is a multiple of 3 2 × False
35 is a factor of 2× 3 3 × False
24 is a factor of 2 4 × 3 2 × True, as 𝟐𝟒= 𝟐 𝟑 ×𝟑; enough 2s and 3s are available within 𝟐 𝟒 × 𝟑 𝟐 ×𝟓. 1 4 ? ? ? ? ? ? ? ? ? ? 2 5 ? ? ? ? ? 6 3 ? ? ? ? ?

29. 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 𝒑 𝟑 , or 𝒑×𝒒, where 𝒑 and 𝒒 are prime numbers. N ?

30. Year 7 HCF/LCM

31. 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, …
Multiples of 12: 12, 24, 36, 48, 60, 72, …
Multiples in common: 24, 48, 72, … ? ? ?

32. Lowest Common Multiple/Highest Common Factor
Multiples of 8: 8, 16, 24, 32, …
Multiples of 12: 12, 24, 36, …
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.

33. Check Your Understanding
Bro Shortcut: Any multiple of 60 ends with a 0. Therefore the multiple of 72 must be x5, x10, … ? 1
𝐿𝐶𝑀 60,72 =𝟑𝟔𝟎 𝐻𝐶𝐹 60,72 =𝟏𝟐 ?
Bro Shortcut: Any number which goes into 60 and 72 must also go into their difference! (i.e. 12) ? 2
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 3 ?

34. But what about bigger numbers?
792, 378
Sometimes it’s not practical to use this method.
Can we use the prime factorisation somehow? ?
792= 2 3 × 3 2 ×11 378=2× 3 3 ×7 ?

35. But what about bigger numbers?
792= 2 3 × 3 2 ×11 378=2× 3 3 ×7
The ‘what wins, what loses’ method
792= 2 3 × ×11 378=2 × 3 3 ×7
Step 1: Align numbers so that each prime factor has its own column.
𝐻𝐶𝐹=2 × 3 2 ?
In terms of the 2s, what do both and 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× =16632 ? ?
This time, what do both and 2 go into? ?
Step 2: To calculate LCM, see ‘what wins’.

36. More Examples
2016= 2 5 × 3 2 ×7 72 = 2 3 × 3 2
588= 2 2 ×3× = 2 4 ×7×11 ? ?
𝑳𝑪𝑴 𝟐𝟎𝟏𝟔,𝟕𝟐 = 𝟐 𝟓 × 𝟑 𝟐 ×𝟕=𝟐𝟎𝟏𝟔 𝑯𝑪𝑭 𝟐𝟎𝟏𝟔,𝟕𝟐 = 𝟐 𝟑 ×𝟑=𝟐𝟒
(note that if there’s a ‘draw’, both win and both lose)
Line numbers up:
𝟓𝟖𝟖 = 𝟐 𝟐 ×𝟑× 𝟕 𝟐 𝟏𝟐𝟑𝟐= 𝟐 𝟒 ×𝟕×𝟏𝟏
𝑳𝑪𝑴 𝟓𝟖𝟖,𝟏𝟐𝟑𝟐 = 𝟐 𝟒 ×𝟑× 𝟕 𝟐 ×𝟏𝟏=𝟐𝟓𝟖𝟕𝟐 𝑯𝑪𝑭 𝟓𝟖𝟖,𝟏𝟐𝟑𝟐 = 𝟐 𝟐 ×𝟕=𝟐𝟖

37. Check Your Understanding?
1936= 𝟐 𝟒 × 𝟏𝟏 𝟐 792= 𝟐 𝟑 × 𝟑 𝟐 ×𝟏𝟏 ?
𝐿𝐶𝑀= 𝟐 𝟒 × 𝟑 𝟐 × 𝟏𝟏 𝟐 =𝟏𝟕𝟒𝟐𝟒 𝐻𝐶𝐹= 𝟐 𝟑 ×𝟏𝟏=𝟖𝟖 ? ?
If you finish… ?
3675=𝟑× 𝟓 𝟐 × 𝟕 𝟐 875= 𝟓 𝟑 ×𝟕 ? ?
𝐿𝐶𝑀=𝟑× 𝟓 𝟑 × 𝟕 𝟐 =𝟏𝟖𝟑𝟕𝟓 𝐻𝐶𝐹= 𝟓 𝟐 ×𝟕=𝟏𝟕𝟓 ?

38. Exercise 4 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Calculator permitted! 5
[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 B C D Solution: B (LCM of 3, 5, 7)
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.
What is 𝜙(10)? 4
What is 𝜙(7)?
What in general is 𝜙(𝑝) for a prime number 𝑝? 𝒑−𝟏
Given that 𝜙 𝑚𝑛 =𝜙 𝑚 𝜙 𝑛 (provided that 𝐻𝐶𝐹 𝑚,𝑛 =1), find 𝜙 =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’.
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
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
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
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 5 1 ? ? ? ? ? 2 6 ? 3 ? ? N ? ? ? ? ? ? ? ? 4 ? ?

39. Year 7 Divisibility

40. Divisibility Rules
How can we tell if a number is divisible by: ! 2
Last digit is even. 3
Digits add up to multiple of 3. e.g: 1692: = 18  4
Last two digits are divisible by 4. e.g 5
Last digit is 0 or 5. 6
Number is divisible by 2 and 3 (so use tests for 2 and 3). 7
Double the last digit and subtract it from the remaining number, and see if the result is divisible by 7.
e.g: > 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 : ( ) – (7 + 4) = 22 – 11 = 11, which is divisible by 11. 12
Number divisible by 3 and by 4. ? ? ? ? ? ? ? ? ? ? ?

41. Quickfire Divisibility 4 6 7 9 11 726  168 9196 252 1001 91 216 87912? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

42. 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 7 11
Is it prime? 91  No
101
Yes
234567
131
781
751
221
No! (13×17) ? ? ? ? ? ? ? ? ? ?
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. ?

43. Test Your Understanding
Easier One:
[JMO 1997 A5] Precisely, one of the numbers 234, 2345, 23456, , , is a prime number. Which one must it be? Solution: ?
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 Solution: A ?

44. 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

45. Quickfire ? ? ? ? ? 18 2 and 9 rules 45 5 and 9 36 4 and 9 40 5 and 8
What divisibility rules would we use if we wanted to test divisibility by: 18
2 and 9 rules 45
5 and 9 36
4 and 9 40
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 is the answer.

46. Exercise 5 Problem sheet of Junior and Intermediate Olympiad problems.
Work in pairs/groups if you wish.
Answers on next slides.
(See printout)

47. 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? ?

48. 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? ?

49. 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? ? ?

50. Question 4
[M55] A palindromic number is one which reads the same when its digits are reversed, for example What is the largest six-digit palindromic number which is exactly divisible by 15? ?

51. Question 5
[J16] Find a rule which predicts exactly when five consecutive integers have sum divisible by 15. ?

52. 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. ?

53. 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. ?

54. Question 8
[M31] Find the smallest positive multiple of 35 whose digits are all the same as each other. ?

55. 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:
Each of 5, 6, 7, 8 is used exactly once in each of the four gaps.
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 = 44 possible numbers divisible by 396.
However, there are total possible ways of using 5, 6, 7 or 8 for each of the four gaps.
The probability is therefore = 11 64

56. N BONUS QUESTION ? Show that:
𝑛 3 −𝑛 is divisible by 6 for all integers 𝑛.
𝑛 4 +2 𝑛 3 − 𝑛 2 −2𝑛 is divisible by 24 for all integers 𝑛. ?
𝑛 3 −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.
= 𝑛 3 𝑛+2 −𝑛 𝑛+2 = 𝑛 3 −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.

57. (any slides after this are supplementary)
Fini
(any slides after this are supplementary)

58. Year 7 Divisibility Within Equations

59. 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
Always false
Sometimes true
𝟓𝒏 is divisible by 5. 
(5 times a number is clearly in the 5 times table)
𝟒𝒏+𝟏 is divisible by 4.
(This will always be one more than a multiple of 4)
𝟒 𝒏+𝟏 is divisible by 4.
𝟑𝒏 is divisible by 2.
𝟔𝒏 is divisible by 2.
N 𝒏 𝒏+𝟏 is divisible by 2.
(one of 𝑛 and 𝑛+1 is even, and
𝑜𝑑𝑑×𝑒𝑣𝑒𝑛=𝑒𝑣𝑒𝑛)
N 𝟒 𝒏 𝟐 is a square
(The product of two squares is a square) ? ? ? ? ? ? ?

60. Since 𝟓𝒙 and 150 are divisible by 5, 𝟒𝒚 must be.
Factors within equations
What can you tell about the following numbers in these equations?
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3+𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3=𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒆 𝒐𝒇 𝟑 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
Since 𝟓𝒙 and 150 are divisible by 5, 𝟒𝒚 must be. ?
What therefore can you say about 𝑦?
4 is not divisible by 5. Therefore 𝒚 must be. ?

61. Divisibility in Equations
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.
Find positive integer solutions to:
5𝑥+4𝑦=50
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
If 𝒚 was 15 then 𝟒𝒚=𝟔𝟎, then 𝒙 would have to be negative. ? ? ?
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 , … ?

62. More examples 7𝑥+6𝑦=90 ? 2𝑥+4𝑦+1=6𝑇 ?
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.

63. 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 C 6 D 7 E 8 𝟖𝒙+𝟓𝒚=𝟑𝟔 𝟓𝒚 is divisible by 4 therefore 𝒚 is. When 𝒚=𝟒, 𝒙=𝟐. So answer is C ?
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 is the only possibility.
If 𝒙=𝟏𝟑, 𝒚=𝟐. Therefore 𝒙+𝒚=𝟏𝟓 ?

64. Exercise 6
[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) 4 𝑥 2 ? Divisible by 36
c) 4 𝑥+1 ? Divisible by 4
c) 2 𝑥 2 3𝑥+1 ? Divisible by 18 3 1
Find all positive integer solutions to the following:
a) 2𝑥+3𝑦= 𝒙,𝒚 = 𝟐,𝟐
b) 4𝑥+5𝑦= 𝒙,𝒚 =(𝟓,𝟑)
c) 7𝑥+5𝑦= 𝒙,𝒚 = 𝟔,𝟕 , (𝟏,𝟏𝟒)
d) 17𝑥+5𝑦= 𝒙,𝒚 = 𝟏𝟓,𝟏𝟕 , 𝟏𝟎,𝟑𝟒 ,(𝟓,𝟓𝟏)
[JMC 2005 Q16] ‘Saturn’ chocolate bars are packed either in boxes of 5 or boxes of 12. What is the smallest number of full boxes required to pack exactly 2005 ‘Saturn’ bars? A B C 168
D E Solution: D ? ? ? 4 ? ? 2 ? 5 ? ? ? ? ?

65. Year 7 Square and cube numbers

66. Square/Cube Numbers
Find the prime factorisation of the following square numbers. What do you notice?
49= 𝟕 𝟐 324= 𝟐 𝟐 × 𝟑 𝟒 = 𝟑 𝟔 × 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= 𝟐 𝟑 8000= 𝟐 𝟔 × 𝟓 𝟑 = 𝟑 𝟗 ? ? ?
!Cube numbers have powers which are multiples of 3.

67. Square/Cube Numbers Q
What is the smallest multiple of 504 that is square?
(Note that 504= 2 3 × 3 2 ×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. ?
Check your understanding:
What do you need to multiply the following by to make the following square and cube? a
Square
Cube
504= 2 3 × 3 2 ×7
×14
×3× (147)
12= 2 2 ×3 ×3
×2× (18)
800= 2 5 × 5 2 ×2
×2×5 (10) ? ? ? ? ?
Bro Reminder: Factors of will be all powers of 2 with a power up to 10. b
How many square numbers are factors of ?
All even powers of 2 ( 𝟐 𝟐 , 𝟐 𝟒 , 𝟐 𝟔 , 𝟐 𝟖 , 𝟐 𝟏𝟎 ). But don’t forget that 𝟏 (= 𝟐 𝟎 ) is square! (as the 0 is even). That’s 6 square factors. ?

68. 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.
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.

69. Exercise 7 (on provided sheet)
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× (a) 6 (b) 180
(ii) 24= 2 3 ×3 (a) 6 (b) 9
(ii) (a) 2 (b) 45
List the cube numbers (in prime factorised form) that are factors of
Solution: 1, 𝟓 𝟑 , 𝟓 𝟔 , 𝟓 𝟗 , 𝟓 𝟏𝟐
[IMC 1999 Q16] Three statements: i) is even (ii) is odd (iii) 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
[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) 1
[Junior Kangaroo 2015 Q23] How many three-digit numbers have an odd number of factors? A B C D 21 E 22
E (22 squares between 100 and 961)
[Cayley 2013 Q1] What is the smallest non-zero multiple of 2, 4, 7 and 8 which is a square?
Solution: 784
[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= 𝟑 𝟐 ×𝟓, then 𝒙 could be 5 to give 𝟑 𝟐 × 𝟓 𝟐 which is square. Thus Anastasia’s number was 5. 6 ? ? ? N1 ? ? 2 N2 ? 3 ? ? 4 ? 5 ?

70. “𝟓𝒙 is divisible by 15”. Supplementary Slides on Divisibility
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. ?

71. Click for Bromanimation
Thinking in buckets
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.
“3𝑥 is divisible by 2” 3 𝑥
The 3 bucket can’t have the factor of 2, so it must have been in the 𝑥 bucket.
Therefore 𝑥 is divisible by 2. 2
Click for Bromanimation
Factors

72. Thinking in buckets 3 “6𝑥 is divisible by 3” 6 𝑥
The 3 may have come from the 6 bucket (6 is divisible by 3). 6 𝑥
But it may or may not have come from the 𝑥 bucket (depending on whether 𝑥 is divisible by 3)
Click for Bropossibility 1
Because the prime factor of 3 could have just come from the 6, we can’t guarantee anything about the divisibility of 𝑥. 3
Factors
Click for Bropossibility 2

73. Thinking in buckets 2 3 “3𝑥 is divisible by 6” ? ? 3 𝑥
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 𝑥
But the 3 might have also been in the 𝑥 bucket. 𝑥 would be divisible by 6.
Click for Bropossibility 1
There what can we guarantee about 𝑥?
It must be divisible by 2, but we can’t guarantee it will be divisible by 6 (because of the first case). ? 2 3 ?
Factors
Click for Bropossibility 2

74. Click for Bromanimation
Thinking in buckets
“ 𝑥 2 is divisible by 3”
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. 𝑥 𝑥 3 3
Click for Bromanimation
Factors

75. We don’t know anything about 𝑥
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.
𝑥 is divisible by 4.
10𝑥 is divisible by 15.
𝑥 is divisible by 5.
𝑥 2 is divisible by 5
𝑥 is divisible by 6.
2 𝑥 2 is divisible by 4
𝑥 is divisible by 25.
2 𝑥 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

76. 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
4𝑥+2
If we add a multiple of 4 and a multiple of 2, it will give a multiple of 2 but not of 4. 2 2 2
Factors
Answer: 8 ?

77. More Examples
If 𝑛 is divisible by 3, what can we say about the divisibility of…
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.
𝑛 3 6𝑛+2
We get a factor of 3 three times from 𝒏 𝟑 . 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. a ?
Test Your Understanding: b ?

78. Supplementary Questions (for teacher use)
[JMO Mentoring Jun2011 Q2] How many positive divisors does 6! have including 6! and 1? [6!=6×5×4×3×2×1.] Solution: 30
[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) 6 1
What are the factors of that are both square and cube? Leave your answer in factorised index form.
Solution: 1, 𝟓 𝟔 , 𝟓 𝟏𝟐 , 𝟓 𝟏𝟖 , 𝟓 𝟐𝟒
How many positive factors does 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
[SMC 2003 Q15] The number of this year, 2003, is prime. How many square numbers are factors of ?
Solution: 1002
[Kangaroo Pink 2010 Q18] How many integers 𝑛, between 1 and 100 inclusive, have the property that 𝑛 𝑛 is a square number? A B C D 10 E 5
Solution: B ? ? 2 7 ? 3 ? ? 4 ? 5 ?

79. BONUS LESSON! Use #3: Trailing zeroes
Prime Factorisation
Number
Trailing Zeroes
2 3 ×3× 5 4
15 000 3
2 5 × 5 2 ×7
5600 2
2 7 × 5 5 × 11 2 5
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. ?

80. Examples ? ? ? How many zeroes are on the end of 2 100 × 3 40 × 5 80 ?
80 trailing zeroes.
8! is said “8 factorial” and means 8!=8×7×…×2×1
How many trailing zeroes does 8! have?
𝟖!= 𝟐 𝟑 ×𝟕×𝟐×𝟑×𝟓× 𝟐 𝟐 ×𝟑×𝟐 = 𝟐 𝟕 × 𝟑 𝟐 ×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 2 7 × 3 2 × 5 5 ?
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
𝟐 𝟐 × 𝟑 𝟐 . This is 36, so the last non-zero digit is 6. Q ? Q ? Q ?

81. Test Your Understanding
How many zeroes are on the end of × 5 90 × ?
40 trailing zeroes.
What is the last non-zero digit of × 5 79 ×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. a ? b ? N ?

82. Exercise 8 (on provided sheet)
How many zeroes are at the end of × 5 31 ? Solution: 31
How many zeroes are at the end of × 3 30 × ?
Solution: 20
What is the last non-zero digit of:
2 10 × 3 2 × Solution: 5
2 8 × 5 7 × 7 2 Solution: 8
2 10 × 5 8 ×13 Solution: 2
[JMO 2010 A3] Tom correctly works out 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 3 4 × 4 5 × 5 6 is written out in full. How many zeroes are there at the end of the number?
Solution: 6
[Kangaroo Pink 2012 Q16] What is the last non-zero digit when × 3 4 × is evaluated?
Solution: 6
[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
[SMC 2001 Q15] Sam correctly calculates the value of 5 8 × How many digits does her answer contain?
Solution: 11
[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
Find two integers, neither of which has a zero digit, whose product is
𝟑𝟐×𝟑𝟏𝟐𝟓 1 7 ? 2 ? ? 8 3 ? ? ? 4 ? 9 ? 5 ? 10 ? 6 ? 11 ? ?