Prime Factorizations, continued
Knowing the multiplication facts for the numbers 1
to 15 significantly helps finding the prime factorizations of larger
numbers.
Example: 225 = 15 * 15 = 3 * 5 * 3 *
5 = 32 * 52
Example: 157 we know 11 *
11 = 121, and 13 * 13 = 169, so
all we need to check are the primes up to 11.
2 does not divide 7,
1+5+7 = 13, 3 does not divide 13, 5 does not divide
7, 7 does not divide
157, we're done, 157 is a prime number.
Greatest Common
Divisor
The greatest common divisor (gcd)
is the largest number that divides two or more integers. (Another name
for the gcd is greatest common factor, abbreviated, gcf.)
Procedure:
to find the gcd
1. write each number in their prime factorizations above one another
2. select the smallest common prime from all numbers, that is each one
with the smallest exponent.
Example:
6, 10
6 = 2 * 3
10 = 2 * 5
gcd = 2
Example: 20,
30, 50
20 = 22 * 5
30
= 2 * 3 * 5
50 = 2 * 52
gcd = 2 * 5 = 10
Example: 8,
15
8 = 23
15
= 3 * 5
gcd = 1
note: no common factor, so we say 8 and 15 are relatively prime.
Word Problems that use
gcd.
Example: A florist has
36 roses, 27 tulips, and 18 carnations she must use to create bouquets. What is
the largest number of bouquets she can make without having any flowers left
over?
Answer: the gcd is the
number we're looking for, the remaining factors in each number tell us how
may of each to put in each bouquet.
roses:
36 = 2 * 18 = 2 * 2 * 9 = 2 * 2 * 3 * 3 = 22 * 32
tulips:
27 = 3 * 9 = 3 * 3 *
3
= 33
carnations: 18
= 2 * 9 = 2 * 3 *
3
= 2 * 32
gcd = 32
= 9 bouquets (smallest exponent of 3 is 2)
each bouquet will
have roses: 22 =
4 (since 4 * 9 = 36)
tulips:
3
(since 3 * 9 = 36)
carnations:
2
(since 2 * 9 = 18)
Example: Say you have
60 pencils, 90 pens and 120 tablets and you want to make packages of pencils,
pens and tablets to donate to your school for students who cannot afford these
supplies. What is the maximum number of packages you can make uisng all
items, and how many pencils, pens and tablets will be in each package?
Answer: This
is another gcd problem, so we have
pencils: 22 * 3
* 5
pens: 2 *
32 * 5
tablets: 23 * 3
* 5
gcd = 2 *
3 * 5 = 30 packages
with 2 pencils, 3 pens, and 4 tablets in each package.
Example:
(advanced) You want to make two garden plots next to each other with a
fence completely around each one. One plot is 180 square feet and
the other is 204 square feet. If the fence comes in 1 foot lengths,
what is the greatest length of the fence you can make that is shared by both
garden plots? How much fencing is required?
Answer: A diagram
might help here.
-------------------------------
| 180
| 204 |
| | |
<-- width
-------------------------------
<-- length
Since area is length * width
we're looking for the greatest common width between these two plots.
180
= 22 * 32 * 5
204 = 22 * 3 *
17
gcd = 22 * 3 = 12
Themaximum shared width is
12 feet
the length of the first plot
will be 3 *5 = 15 feet since 15 * 12 = 180
and the length of the second
plot will be 17 feet since 17 * 12 = 204
So we have a 12 * 15
plot next to a 12 * 17 plot.
The fencing required is
12 + 15 + 12 + 15 = 24 + 30 = 54 feet for the first plot
and 17 + 12 + 17 = 12 +
34 = 36 feet for the second plot
(remember the first plot
contains the common fence)
So we need 54 + 36 = 90
feet of fence.
Back to Integers
Prime Numbers
Least Common Multiple
This is another
FREE ALGEBRA PRINTABLE presented to you from the
Algebra section of
K12math.com