Download Factors, Prime Factorization, Common Factors

4.2 Factors and Divisibility
Numbers that are multiplied together are called factors.
Factors of a number, a, are numbers that when multiplied
together produce a product of a.
The number 12 has 6 possible factors:
1 x 12 = 12
2 x 6 = 12
3 x 4 = 12
So the factors are 1, 2, 3, 4, 6, and 12.
A factor pair table can be written for 12 to find all the factors.
Just start from 1 and keep incrementing and checking if 12 is divisible by
it. When a factor you choose has already appeared in the table. You
know you are finished.
12
Note that a number is always a factor of itself because a x 1 = a
When two whole numbers, such as 3 and 2, multiply to get a product, 6
(3 x 2 = 6), then we can say these facts:
1) 6 is a multiple of 3 and 2
2) 6 is divisble by 3 and 2
3) 3 and 2 divide evenly into 6.
4) 3 and 2 are factors of 6.
Example: 12 is a multiple of 4 and a multiple of 3
12 is divisible by 3 and 4
3 and 4 divide evenly into 12
3 and 4 are factors of 12.
Division Review
5
10
10  2  10 / 2   5, also we can do it this way : 2 10
2
Dividend
Divisor
Quotient
Divisor
Quotient
Dividend
Division Properties:
Division is the “inverse” of multiplication. That is, does the opposite of what
multiplying does.
If 4 times 5 is 20, then 20÷5 answers the question,
4(5) = 20
“5 times what equals 20?”
4 = 20÷5
and 20÷4 answers the question,
“4 times what equals 20?”
5 = 20÷4
Division with 0
0 divided by any nonzero number is 0. 0÷5 = 0
Any nonzero number divided by 0 is undefined.
5÷0 is undefined.
0/0 is undetermined.
More Division Properties
Any number divided by 1 is itself. 5÷1 = 5
Any number divided by itself is 1 (provided that number ≠0)
5÷5 = 1
When dividing by multiples of 10, you may cancel an equal
number of “right-hand” zeroes.
600÷10=60
8,000÷200 = 80÷2 = 40
32
3 96
-9
6
-6
0
No remainder.
96 is exactly
divisible by 3.
96÷3 = 32
Check:
32 x 3 = 96
36
4 147
-12
27
-24
3
258
7 1809
-14
40
-35
59
-56
3
There are no more digits
left to carry down, so 3 is
the remainder and 147 is
not exactly divisible by 4.
147÷4 = 36 R3
“36 with 3 left over”
There are no more digits
left to carry down, so 3 is
the remainder and 1809 is
not exactly divisible by 7.
147÷7 = 36 R3
“36 with 3 left over”
Check:
36 x 4 + 3 = 147
Don’t forget to add the
remainder if there is
one.
Check:
258 x 7 + 3 = 1809
Don’t forget to add the
remainder if there is
one.
Checking Division:
Quotient X Divisor + Remainder = Dividend
Translation: The answer to the division problem times what you divided by, plus the remainder,
should equal the number you divided into.
Div.
By
Divisibility Tests
Example
2
A number is divisible by 2 if the last digit is 0, 2, 4, 6 or 8.
168 is divisible by 2 since
the last digit is 8.
3
A number is divisible by 3 if the sum of the digits is divisible
by 3.
168 is divisible by 3 since
the sum of the digits
is 15 (1+6+8=15),
and 15 is divisible
by 3.
4
A number is divisible by 4 if the number formed by the last
two digits is divisible by 4.
(also, a number if divisible by 4 if after being halved, it is still
even).
316 is divisible by 4 since
16 is divisible by 4.
72 is divisible by 4
because half of 72 is
36 and 36 is even.
5
A number is divisible by 5 if the last digit is either 0 or 5.
195 is divisible by 5 since
the last digit is 5.
6
A number is divisible by 6 if it is divisible by 2 AND it is
divisible by 3.
168 is divisible by 6 since
it is divisible by 2
AND it is divisible
by 3.
7
Take the last digit, double it, and subtract it from the rest of
the number; if the answer is divisible by 7 (including 0),
then the number is also.
If you had 203, you
would double the last
digit,3, to get 6, and
subtract that from 20 to
get 14, which is divisible
by 7.
8
A number is divisible by 8 if the number formed by the last
three digits is divisible by 8.
7,120 is divisible by 8
since 120 is
divisible by 8.
9
A number is divisible by 9 if the sum of the digits is divisible
by 9.
549 is divisible by 9 since
the sum of the digits
is 18 (5+4+9=18),
and 18 is divisible
by 9.
10
A number is divisible by 10 if the last digit is 0.
1,470 is divisible by 10
since the last digit is
0.
Example
A total of 216 girls tried out for a city volleyball program. How many girls
should be put on the team roster if the following requirements must be
met?
1)All the teams must have the same number of players.
(find a number that goes exactly into 216, so there is no remainder)
2) A reasonable number of players on a team is 7 to 10 (divide 216 players by
7 players per team, then 8, then 9, then 10). But don’t bother with 10
because we know 10 doesn’t go exactly into 216.
3) There must be an even number of teams. (The quotient must be EVEN).
6<7, and there
are no more digits
left to carry down,
so 6 is the
remainder and
216 is not exactly
divisible by 7.
3
7 216
21
6
27
8 216
16
56
56
0
8 goes exactly
into 216, but the
quotient is 27,
which is ODD.
(Does not meet
3rd req.)
24
9 216
18
36
36
0
This meets all 3
requirements:
9 players per team leaves no
remainder, 9 is an acceptable
number for a team (which is
a number between 7 and 10),
and the number of teams is
24, which is an even number.
Prime Numbers
A prime number is a whole number, greater than 1, that has only
1 an itself as factors.
Composite Numbers
A composite number is a whole number, greater than 1, that are
not prime.
Prime Factorization
To find the prime factorization of a whole number means to write it as
the product of only prime numbers.
This is can be useful when finding things like the Greatest Common Factor or
Least Common Multiple between two numbers (we’ll get into that later).
Example: Factor 90 into its prime factors.
Choose any two
factors of 90
(besides 1 and
90)
Then do the
same with each
of those factors.
Keep going until
you have only
prime factors as
the bottom “roots”
of the “factor
tree.”
90
9
3
10
3
2
5
90 = 3 3 2 5
●
●
●
Putting these factors in numerical order and then
re-writing repeated factors as powers gives:
90 = 2●32●5
Theorem:
Any composite number has exactly one set of prime factors.
Example 5
Find the prime factorization of 210
First, pick any two factors of 210.
For instance 21 and 10.
We could have also picked 7 and 30 as the
factors.
210
21
210
10
7
30d
d
3
7
2
5
Notice that either method gives us
210 = 2●3●d5●7
d
6
3
d
2
5
Alternate Factoring Method –
I call it the “Pyramid Division Method”
For more info:
http://www.purplemath.com/modules/factnumb.htm
With this method, you start and the bottom of the pyramid
and move up, so you have to leave lots of room at the top
of your problem.
Example: Find the prime factorization of 90.
Start with the first prime factor you can think of that goes
into 90. We can’t choose 9 or 10 because they aren’t
prime. Since 90 is even, start with the factor 2.
3
39
5 45
2
is 3 prime? Yes! You are done.
is 9 prime? No. Keep dividing.
is 45 prime? No. Keep dividing.
90
Prime Factorization of 90 is 2 x 3 x 3 x 5
Write the factors in numerical order and use exponents where factors are repeated
=
 2  32  5
Upside-Down Division Method
Example: Find the prime factorization of 60
2
60
3
30 Is 30 prime? No.
2
10 Is 10 prime? No.
5 Is 5 prime? Yes. Stop
Prime Factorization of 60 is
22  3  5
The Greatest Common Factor is the largest number that goes
into two numbers.
What is the GCF of 20 and 24? The largest number that goes
into both 20 and 24 (the largest divisor of both 20 and 24)
Both numbers are divisible by 2. Is 2 the GCF?
20÷2 = 10, and 24÷2 = 12. But 10 and 12 have common factors,
so 2 is not the GCF of 20 and 24.
A bigger number can go into both 20 and 24.
4 is the GCF of 20 and 24. 20÷4 = 5, and 24÷4 = 6.
5 and 6 have no common factors, so 4 is the GCF.
Finding the GCF with Prime Factorization
Find the GCF of 24 and 32
24
32
24
32
4
2
6
2
2
4
3
2
24= 2● 2 ● 2 ● 3
8
2
2
4
2
2
32 = 2● 2 ● 2● 2 ● 2
24  2  2  2  3
32  2  2  2  2  2
Cancel out factors that are in COMMON in both
factorizations. The factors that you crossed out make up
the GCF.
GCF =
2 2 2  8