Number Operations

$GCSE Maths Revision$

Here we shall go through various number operations. In this article we’ll assume that you know the basics such as multiplication and division.

Prime Factor Decomposition

Whilst this sounds a little intimidating, if we break it down we can see that it is actually very simple. Prime factors are simply all the factors of a number that are prime numbers. To decompose something means to break it down, so the Prime Factor Decomposition of a number is simply the number broken down into it’s prime factors, let’s have an example.

Find the prime factor decomposition of 54.

To tackle this problem we take a look at 54 and see what numbers divide it. As 54 is even we know it can be divided by 2.

54 = 2 x 27

2 is a prime number, there we cannot divide this anymore. However 27 is not. Notice that you can divide 27 by 3.

27 = 3 x 9

Nearly there! However 9 is still not a prime number, but it can be divided by 3.

9 = 3 x 3

We have found all the prime factors of 54.

54 = 2 x 3 x 3 or 54 = $2 \times 3^2$

It is important to remember that any number can be written as the product of it’s prime factors.

LCM

The lowest common multiple (LCM) of a set of numbers is the smallest integer that they divide into without leaving any remainders. For example, the LCM of 3, 5, 6 and 15 would be 30.

HCF

The highest common factor (HCF) of a set of numbers is the largest number that is a factor of the given numbers. For example, the HCF of 24 and 16 is 8.

Approximations

There are times in the real world when approximations are needed to simplify certain measurements and calculations. For example, if you ask someone what is the population of the UK, no one would take a guess at 61,399,118, whereas most people would guess at about 60 million. The point is that in real life, approximations can often be a lot easier to deal with than the actual value.

However it may be the case that we only get given the approximations, in cases such as these you are often asked to calculate the upper and lower bounds of the true value. Let’s take a look at a couple of examples.

The attendance at a Saturday afternoon football match is 75,000 to the nearest 500 people. Find the upper and lower limits of this value.

We are told in the question that the given figure is rounded to the nearest 500 people. To find our two values we add and subtract 500 from the given number. The lower and upper bounds of the attendance is therefore 74,500-75,500.

Another example of this type of question involve calculating the upper and lower bounds of areas. For example:

A picture frame is 90cm by 150 cm to the nearest 5 cm. State the range of values the area can take.

Although this sounds a little intimidating it’s pretty much the same as the last example.

We can work out in the same way as above that the limits of the two lengths are 85-95 and 145-155 respectively. The question asked for the range of values of the area, to find this we multiply the two shortest lengths together, and the two longest lengths together.

85 x 145 = 12325cm² and 95 x 155 = 14725cm²

The range of values that the area can take is 12325cm² – 14725cm²

BODMAS/BIDMAS

When we have larger expressions, it can often be confusing what operations do do first, for example.

Find the value of 5 + 7 x 3 – 2(7 – 3)

We could work this out in a number of ways:

5 + (7 x 3) – 2(7 – 3) = 5 + 21 – 8 = 18

or

(5 + 7) x 3 – 2(7 – 3) = 36 – 8 = 28

Which one is correct?

It is in situations like this that we use BODMAS. This is simply the order in which we carry out various operations, with each of the letters standing for a particular operation.

B – Brackets – Calculate the expressions contained in brackets first.
O – Order – Calculate terms such as $2 \times 3^2$ and $\sqrt{2}$ second.
D – Division – Calculate the terms involving division next.
M – Multiplication – Next comes the terms involving multiplication.
A – Addition – Now we work out the terms involving addition
S – Finally it’s the turn of terms involving subtraction.