Download The Mathematics 11 Competency Test

The Mathematics 11
Competency Test
Prime Factors
To factor a whole number is to write it as a product of two or more other whole numbers. The
individual numbers in this product are called factors of the original number.
A prime number or prime is a whole number which is evenly divisible only by itself and 1. In
other words, prime numbers have only 1 and themselves as factors. Whether or not the number
1 should be considered a prime number is a matter of futile debate. Ignoring 1, the first few prime
numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
101, 103, ….
There is no known formula which generates all prime numbers. Instead, to establish that a
number is prime, you have to use a method of systematic trial and elimination of potential factors
to demonstrate that it has no factors other than itself and 1. Mathematicians have been intrigued
by prime numbers for thousands of years, and continue to study their properties enthusiastically
today because they have applications in many problems of technology.
In simplifying fractions, it is useful to begin by factoring the numerator and denominator into a
product of factors which are all prime numbers – hence the term prime factor. (We also did this
earlier in these notes in attempting to simplify square roots of numbers.) The procedure is very
straightforward and systematic:


start by checking for factors of 2 in the number, and remove as many as possible
then check for factors of 3 in what’s left after all factors of 2 have been removed, and
remove as many factors of 3 as possible

repeat this process on what’s left after previous prime factors have been removed, using
the prime numbers in turn: 5, 7, 11, 13, … etc. When it is clear that what’s left to be
factored is a prime number itself (and so cannot be factored further), the process is done.
Example: Find the prime factors of 156.
Solution:
 156 is even, so it contains a factor of 2:
156  2 

156
 2  78
2
78 is even, so it contains a factor of 2:
78 

156  2   2 
 2  2  39
2 


39 is not even, so there are no further prime factors of 2. So, check if 39 is divisible by 3.
Since
39  13 , a whole number, we conclude that 3 is a prime factor of 39, and so
3
David W. Sabo (2003)
Prime Factors
Page 1 of 3
39 

156  2  2   3 
 2  2  3  13
3 

Having removed three prime factors ( 2, 2, and 3) from 156, we are left with the factor 13. But 13
is itself a prime number. Thus 156, written as a product of prime factors is
156  2  2  3  13
Notice that as we remove prime factors, the number on which we need to focus further attention
is always getting smaller and smaller.
We’ll do one more really long example to make sure you clearly understand this systematic
approach to determining the prime factors of a number. Usually however, the numbers we need
to factor are much smaller (hence the factoring process is much shorter and less tedious) than
the one in the next example.
Example: Find the prime factors of 58212.
Solution:

58212 is even, so it contains a factor of 2:
58212  2 

58212
 2  29106
2
29106 is even, so it contains a factor of 2:
29106 

58212  2   2 
 2  2  14553
2 


14553 is not even, so it does not contain a factor of 2. But, since 3 divides evenly into
14553, we have that there is a factor of 3 here:
14553 

58212  2  2   3 
 2  2  3  4851
3 


we are now working on factors of 3. Again, 3 divides evenly into 4851, so there is at least
one more factor of 3:
4851

58212  2  2  3   3 
 2  2  3  3  1617
3 


we are still working on factors of 3. Again, 3 divides evenly into 1617, so there is at least
one more factor of 3:
1617 

58212  2  2  3  3   3 
 2  2  3  3  3  539
3 

David W. Sabo (2003)
Prime Factors
Page 2 of 3

we are still working on factors of 3. However, a test calculation shows that 3 does not
divide evenly into 539, so we conclude there are no more factors of 3 in the original
number. Clearly 539 is not divisible by 5. So the next prime number to try is 7. We find
that 539 is evenly divisible by 7, giving
539 

58212  2  2  3  3  3   7 
 2  2  3  3  3  7  77
7 


We can see that the remaining unfactored part, the 77, is still evenly divisible by 7, so we
have:
77 

58212  2  2  3  3  3  7   7 
 2  2  3  3  3  7  7  11
7 

This is the end, because all of the numbers in the product on the right are prime numbers. Thus,
the prime factorization of 58212 is:
58212  2  2  3  3  3  7  7  11
You can easily verify that multiplying the numbers on the right together does give 58212, and all
seven factors shown are prime numbers. Thus, this is the requested solution to the problem.
David W. Sabo (2003)
Prime Factors
Page 3 of 3