Calculate Highest Common Factors(HCFs)
& Least Common Multiples(LCMs) – NA1
What are the multiples of 5?
Hey Look…
The multiples
are in the five
times table
Don’t forget…
1
2
3
4
×
×
×
×
5
5
5
5
1
2
3
5
6
9
×
×
×
×
×
×
90
45
30
18
15
10
=
=
=
=
the number itself is a multiple.
5
10
15
20
So 5, 10, 15, 20, … etc are multiples of 5
What are the factors of 90?
Each of these is a pair of factors.
There are 6 pairs of
factors, hence 12 factors.
=
=
=
=
=
=
Don’t forget…
90
90
90
90
90
90
1 and the number itself
are also factors.
So 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
are factors of 90
Lowest Common Multiples
Strategy
List the multiples of the two
numbers & ask 'what is the
smallest number common
to both lists?'
This is the LCM.
Example
Calculate the Lowest Common Multiple of 45 and 60.
Solution
Multiples of 45: 45, 90, 135, 180, 225, …
Multiples of 60: 60, 120, 180, 240, …
It’s 180
So 180 is the LCM of 45 and 60
Your Turn!!
a) Calculate the Lowest Common Multiple of 8 and 12.
Definition
The Lowest Common Multiple of 2 or more numbers is the lowest number that can be divided by all of these numbers.
Strategy
List the factors of the two
numbers & ask 'what is the
largest number common to
both lists?'
This is the HCF.
Highest Common Factors
Example
Calculate the Highest Common Factor of 45 and 60.
Solution
Factors of 45: 1, 3, 5, 9, 15, 45.
Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
It’s 15
So 15 is the HCF of 45 and 60
Your Turn!!
b) Calculate the Highest Common Factor of 8 and 12.
Definition
The Highest Common Factor of 2 or more numbers is the highest number that can divide into all of these numbers.
Observation
180 is the Lowest Common Multiple of 45 and 60.
15 is the Highest Common Factor of 45 and 60.
The Lowest Common Multiple is a ‘BIG’ number
The Highest Common Factor is a ‘SMALL’ number
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
1.
Calculate the LCM of 12 and 20.
2.
Calculate the HCF of 12 and 20.
© ZigZag Education, 2004
Know what a Prime Number is – NA2
A prime number is special because it only has TWO factors: 1 and itself.
Here is a list of factors for the first few whole numbers.
1:
2:
3:
4:
5:
6:
7:
1
1,
1,
1,
1,
1,
1,
8
2.
9
3.
9
2, 4. 8
5.
9
2, 3, 6. 8
7.
9
Prime Numbers
Which numbers have exactly TWO factors?
The first seven prime numbers are: 2, 3, 5, 7, 11, 13, 17,…
Don’t forget…
1 is not a prime number as it has only ONE factor
Strategy
Look for the factors systematically by dividing
the number using the prime numbers
Example
Are either of 101 or 1001 prime?
Solution
Ask the question
Try the next prime
Does
Does
Does
Does
Does
2 go into 101 exactly? 8
3 go into 101 exactly? 8
5 go into 101 exactly? 8
7 go into 101 exactly? 8
13 go into 101 exactly? 8
Does
Does
Does
Does
2
3
5
7
go
go
go
go
into
into
into
into
1001
1001
1001
1001
exactly?
exactly?
exactly?
exactly?
8
8
8
9
If you find another factor stop!
You can keep trying but you will find no additional factors!
1001 is NOT prime because it has at
least three factors i.e. 1, 1001 and 7!
101 is prime.
Your Turn!!
a) Blank out the above before you start; now list the numbers 1 to 25 and decide which are prime.
Extra
When do you stop trying to find factors?
Strategy
Stop looking when you have tried all of the prime numbers up to the square root of the number.
Example
Continuing from the example of whether 101 is a prime number,
13 × 13 is bigger than 101 so there is no factor between 1 and 101
There is no additional factor between 1 and
i.e. 10.05.
101 , so there are no additional factors.
If there is an additional factor pair, then the smallest factor
of the pair must be between 1 and the square root of the number.
Hey…
Why stop there?
It is based on the idea
that factors come in
pairs AND (in this
example) that the
missing factor pair lies
between 1 × 101 and
10.05 × 10.05.
Write Whole Numbers as the Product of Primes – NA2
Example
Write 90 as the product of primes.
Solution
Method
Draw a factor tree
90 does divide by 2 exactly!
90
2
45
Product means multiplication
45 does not divide by 2, so try 3
15
3
Remember…
Strategy
Begin by dividing by the first prime number, 2.
If the number does not divide by this prime, then try
dividing by the next prime and then the next etc.
3
5
So 90 written as a product of primes: 90 = 2 × 3 × 3 × 5
Hey Look…
This is 90 written as its
product of prime factors
Your Turn!!
b) Write 1260 as the product of primes.
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
1. What is a prime number?
2. Which of 2, 19, 77, 100 are prime?
3. Write 100 as the product of primes.
© ZigZag Education, 2004
Calculate with Negative Numbers with & without a Calculator using + – × ÷ NA3
+
–
–
+
Two negatives multiplied together are positive. Similarly,
two negatives one divided by the other are positive.
The results are summarised as follows:
Example
1. –2 × –4
2. –4 ÷ 2
3. 4 × –2
4. 4 ÷ 2
=
=
=
=
+
–
+
–
=
=
=
=
+
+
–
–
+
–
–
+
÷
÷
÷
÷
+
–
+
–
=
=
=
=
+
+
–
–
Also use the rules when + – symbols are next to each other!
Example
5. 3 – – 3 or written 3 – (–3) = 3 + 3 = 6
6. 3 + – 3 or written 3 + (–3) = 3 – 3 = 0
8
–2
–8
2
Your Turn!!
Find
×
×
×
×
a) –8 × –3
b) –12 × 9
c)
Hint: This is a way
10
−2
d) –8 ÷ –4
of writing10 ÷ –2
e) 5 – – 6
Calculator Check – Make sure you can use your calculator – each one is different
Do the examples 1-4 above on your calculator. Check you get the 4 results shown. To enter a negative number
use the ± button or the (− ) button, e.g. To enter –1 × –6, press, (− ) 1 × (− ) 6 OR 1 ± × 6 ± .
Reminder
Normal subtractions are calculated by simply moving along the number line.
Example
–3 – 4 = –7
5 – 6 = –1
-4
-7
-6
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
Find
1. – 8 × –4
2. 3 – – 12
3. –1 – 4
4.
−10
−5
5.
−4
2
6. 3 – + 12
Substitute Numbers into a Formula – NA4
Example 1
Replace the letter, x, with
its specified value, –2.
For the equation, y = 3x + 7,
find y when x is –2.
It’s an excellent habit to place the
substituted values in brackets!!
Solution
y = 3 × (–2) + 7
y = –6 + 7
y=1
Reminder
+×+
–×−
–×+
+×–
=
=
=
=
+
+
−
–
Look at the sign to the
left of the number to help
you do the multiplication.
Example 2
For the equation, y = 3x2 – 7x + 2,
find y when x is –2
Remember
Solution
(–2) 2 = (–2) × (–2) = 4
y = 3(–2)2 – 7(–2) + 2
y = 3 × 4 + 14 + 2
–7 × (–2) = 14
Your Turn!!
a) In Examples 1) and 2) above, find y when x is 2.
y = 12 + 14 + 2 = 28
b) In Examples 1) and 2) above, find y when x is –3.
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
1. Find y when x is –2, where y = 7 – 3x
2. Find y when x is –5, where y = 1 – 3x2
© ZigZag Education, 2004
Know & use the Index Laws (using numbers) – NA5
Index Laws
30 = 1
3a × 3b = 3a+b
3a ÷ 3b = 3a–b
(3a)b = 3a×b
All numbers to the power 0 are 1
For multiplication just add the indices
For division just subtract the indices
For power (repeated multiplication) just
multiply the indices
These are the index laws
using the number 3.
So 4a × 4b = 4a+b
The laws work with any number as long as the base number is the same!
But watch out with 4 a × 5 b.
The base numbers are different so index laws don’t work!!
Why do the index laws work?
Calculate 32 × 34
32 means 3 × 3
4
3 means 3 × 3 × 3 × 3
So 32 × 34 means 3 × 3 × 3 × 3 × 3 × 3 which is just 36
Hey Look…
5
35 ÷ 32 means
3
3× 3× 3× 3× 3
which means
.
2
3× 3
3
Cancel the 3’s,
3× 3× 3× 3× 3
= 3 × 3 × 3 = 33
3× 3
This is just the two indices added i.e. 2 + 4
Hey Look…
This is just the two indices subtracted i.e. 5 – 2
(35)3 means 35 × 35 × 35, this means (3 × 3 × 3 × 3 × 3) × (3 × 3 × 3 × 3 × 3) × (3 × 3 × 3 × 3 × 3) = 315
We could just add these indices 5 + 5 + 5 = 15!!
Your Turn!!
a) Learn the four index laws above
Hey Look…
This is just the two indices, 5 and 3, multiplied i.e. 5 × 3
b) Write i) 36 ÷ 32 in the form 3a
ii) (44)6 in the form 4b
Know & use the Index Laws (using letters) – NA5
Index laws
a0 = 1
ax × ay = ax+y
Hey Look…
All numbers to the power 0 are 1
For multiplication just add the indices
a
x
a
y
ax ÷ ay = ax–y
For division just subtract the indices – also written
= ax–y
(ax)y = axy
For power (repeated multiplication) just multiply the indices
These laws are essentially
the same as above but
having replaced the base
number 3 with the letter a
Example
a.
a3 × a9 = a3 + 9 = a12
b.
b4 ÷ b3 = b4 – 3 = b1 = b
The letters must be the same
for the index laws to work!
Remember…
c.
4 7
4×7
(c ) = c
28
=c
Your Turn!!
c) Learn these four index laws
Numbers to the power 1 are just themselves e.g. 31 = 3 OR b 1 = b
d) Simplify i) e5 ÷ e2
ii) (m8)4
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
Simplify using the index laws.
1. 37 × 38
2. 38 ÷ 37
3. (38)7
4. x3 × x9
5. y5 ÷ y3
6. (z4)7
© ZigZag Education, 2004
Round to a Given Number of Significant Figures (s.f.) – NA6
ignore these!
Example
3 .1 4 1 6 9 2 6 5 4
Round 3.141692654 to 1 s.f.
Solution
The number after the required significant figure is
So round down.
Round down by ignoring numbers after 3.
So this becomes just 3 to 1 s.f.!!
Example
1.
3 .1 4 1 6 9 2 6 5 4
Round 3.141692654 to 5 s.f.
Solution
This is the 5th significant figure.
The number after this is 9.
This is ‘5 or higher’ so round up.
Round up by increasing this by one.
So this becomes 3.1417 to 5 s.f.!!
Your Turn!!
Round 3.140692654 to a) 2 s.f.
1 s.f. just means the first
digit which isn’t a zero,
i.e. in this case 3.
Strategy
Look at the number after the required significant figure.
Round up if this number is 5 or higher. Round down if
this number is 4 or below. Round down by ignoring
numbers after the required significant figure. Round up
by rounding the required significant figure up by one
digit and ignore numbers thereafter.
6+1 = 7
b) 3 s.f.
c) 4 s.f.
For SMALL numbers between 0 and 1 Æ Watch Out!!
Example
Round 0.34056 to 3 s.f.
Front noughts are not significant
Solution
This is the 3rd significant figure.
The number after this is 5.
This is ‘5 and over’ so round up.
Round up by increasing this by one.
So this becomes 0.341 to 3 s.f. !!
Your Turn!!
d) Round 0.3 4 05 6 to 2 s.f.
Hint: these noughts are not significant.
0.34056
This is not a front nought so it is significant
0+1 = 1
This is a front nought so it is not significant!
e) Round 0. 0 0 3 4 0 5 6 to 4 s.f.
These ones still are significant
Extra notes on Estimation
In your examination you will be expected to round your answers as appropriate, often to 2 or 3 significant
figures. Occasional examination questions will require you to estimate a simple numerical calculation. Normally
this requires you to simply round the numbers to a convenient number and then calculate - often rounding to 1
significant figure is appropriate.
Example 1
Solution 1
Example 2
Solution 2
20.04 + 5.4
20 + 5 25
≈
=
= 12.5
Estimate 2.05 × 8.21 × (3.9)2 ≈ 2 × 8 × (4) 2 = 16 × 16 = 16
Estimate
2
2
2.15
Note: ≈ means approximately equals to
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
Round the following to the number of significant figures in brackets.
1. 70.254 (2)
2. 70.254 (3)
3. 0.05454 (2)
4. 0.05454 (3)
© ZigZag Education, 2004